RADIO-FREQUENCY AND MICROWAVE COMMUNICATION CIRCUITS
RADIO-FREQUENCY AND MICROWAVE COMMUNICATION CIRCUITS Analysis an...

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Devendra K. Misra

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RADIO-FREQUENCY AND MICROWAVE COMMUNICATION CIRCUITS

RADIO-FREQUENCY AND MICROWAVE COMMUNICATION CIRCUITS Analysis and Design Second Edition

Devendra K. Misra University of Wisconsin–Milwaukee

A JOHN WILEY & SONS, INC., PUBLICATION

Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax 978-646-8600, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services please contact our Customer Care Department within the U.S. at 877-762-2974, outside the U.S. at 317-572-3993 or fax 317-572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic format. Library of Congress Cataloging-in-Publication Data: Misra, Devendra, 1949– Radio-frequency and microwave communication circuits : analysis and design / Devendra K. Misra.—2nd ed. p. cm. Includes bibliographical references and index. ISBN 0-471-47873-3 (Cloth) 1. Radar circuits–Design and construction. 2. Microwave circuits–Design and construction. 3. Electronic circuit design. 4. Radio frequency. I. Title. TK6560 .M54 2004 621.384 12–dc22 2003026691 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1

CONTENTS

Preface

ix

1 Introduction 1.1 Microwave Transmission Lines 1.2 Transmitter and Receiver Architectures 2 Communication Systems 2.1 2.2 2.3 2.4 2.5

Terrestrial Communication Satellite Communication Radio-Frequency Wireless Services Antenna Systems Noise and Distortion Suggested Reading Problems

3 Transmission Lines 3.1 3.2 3.3 3.4

Distributed Circuit Analysis of Transmission Lines Sending-End Impedance Standing Wave and Standing Wave Ratio Smith Chart Suggested Reading Problems

1 4 8 11 12 13 16 18 33 52 52 57 57 67 80 85 95 95 v

vi

CONTENTS

4 Electromagnetic Fields and Waves 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

Fundamental Laws of Electromagnetic Fields The Wave Equation and Uniform Plane Wave Solutions Boundary Conditions Uniform Plane Wave Incident Normally on an Interface Modified Maxwell’s Equations and Potential Functions Construction of Solutions Metallic Parallel-Plate Waveguide Metallic Rectangular Waveguide Metallic Circular Waveguide Suggested Reading Problems

5 Resonant Circuits 5.1 5.2 5.3 5.4 5.5

Series Resonant Circuits Parallel Resonant Circuits Transformer-Coupled Circuits Transmission Line Resonant Circuits Microwave Resonators Suggested Reading Problems

6 Impedance-Matching Networks 6.1 Single Reactive Element or Stub Matching Networks 6.2 Double-Stub Matching Networks 6.3 Matching Networks Using Lumped Elements Suggested Reading Problems 7 Impedance Transformers 7.1 Single-Section Quarter-Wave Transformers 7.2 Multisection Quarter-Wave Transformers 7.3 Transformer with Uniformly Distributed Section Reflection Coefficients 7.4 Binomial Transformers 7.5 Chebyshev Transformers 7.6 Exact Formulation and Design of Multisection Impedance Transformers 7.7 Tapered Transmission Lines 7.8 Synthesis of Transmission Line Tapers 7.9 Bode–Fano Constraints for Lossless Matching Networks

104 104 114 119 123 126 130 133 137 142 145 145 151 151 160 164 170 177 184 184 189 190 202 207 226 226 234 234 237 239 244 248 255 263 270 280

CONTENTS

Suggested Reading Problems 8 Two-Port Networks 8.1 8.2 8.3 8.4 8.5

Impedance Parameters Admittance Parameters Hybrid Parameters Transmission Parameters Conversion of Impedance, Admittance, Chain, and Hybrid Parameters 8.6 Scattering Parameters 8.7 Conversion From Impedance, Admittance, Chain, and Hybrid Parameters to Scattering Parameters, or Vice Versa 8.8 Chain Scattering Parameters Suggested Reading Problems 9 Filter Design 9.1 Image Parameter Method 9.2 Insertion-Loss Method 9.3 Microwave Filters Suggested Reading Problems 10 Signal-Flow Graphs and Their Applications 10.1 Definitions and Manipulation of Signal-Flow Graphs 10.2 Signal-Flow Graph Representation of a Voltage Source 10.3 Signal-Flow Graph Representation of a Passive Single-Port Device 10.4 Power Gain Equations Suggested Reading Problems 11 Transistor Amplifier Design 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Stability Considerations Amplifier Design for Maximum Gain Constant-Gain Circles Constant Noise Figure Circles Broadband Amplifiers Small-Signal Equivalent-Circuit Models of Transistors DC Bias Circuits for Transistors

vii

281 281 283 284 289 296 298 304 304 323 325 325 326 333 334 353 380 389 389 392 396 401 402 410 417 418 422 422 429 439 457 466 469 472

viii

CONTENTS

Suggested Reading Problems 12 Oscillator Design 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Feedback and Basic Concepts Crystal Oscillators Electronic Tuning of Oscillators Phase-Locked Loop Frequency Synthesizers One-Port Negative Resistance Oscillators Microwave Transistor Oscillators Suggested Reading Problems

13 Detectors and Mixers 13.1 13.2 13.3 13.4 13.5 13.6

Amplitude Modulation Frequency Modulation Switching-Type Mixers Conversion Loss Intermodulation Distortion in Diode-Ring Mixers FET Mixers Suggested Reading Problems

475 476 479 479 492 494 497 516 520 523 538 538 543 544 555 559 565 567 571 577 577

Appendix 1

Decibels and Neper

580

Appendix 2

Characteristics of Selected Transmission Lines

582

Appendix 3

Specifications of Selected Coaxial Lines and Waveguides

588

Appendix 4

Some Mathematical Formulas

590

Appendix 5

Vector Identities

593

Appendix 6

Some Useful Network Transformations

596

Appendix 7

Properties of Some Materials

599

Appendix 8

Common Abbreviations

601

Appendix 9

Physical Constants

609

Index

611

PREFACE

Wireless technology continues to grow at a tremendous rate, with new applications still reported almost daily. In addition to the traditional applications in communications, such as radio and television, radio-frequency (RF) and microwaves are being used in cordless phones, cellular communication, local area networks, and personal communication systems. Keyless door entry, radio-frequency identification, monitoring of patients in a hospital or a nursing home, cordless mice or keyboards for computers, and wireless networking of home appliances are some of the other areas where RF technology is being employed. Although some of these applications have traditionally used infrared technology, RF circuits are taking over, because of their superior performance. The present rate of growth in RF technology is expected to continue in the foreseeable future. These advances require the addition of personnel in the areas of radio-frequency and microwave engineering. Therefore, in addition to regular courses as a part of electrical engineering curriculums, short courses and workshops are regularly conducted in these areas for practicing engineers. This edition of the book maintains the earlier approach of a presentation based on a basic course in electronic circuits. At the same time, a new chapter on electromagnetic fields has been added, following several constructive suggestions from those who used the first edition. It provides the added option of using the book for a traditional microwave engineering course with electromagnetic fields and waves. Or, this chapter can be bypassed so as to follow the approach used in the first edition: that is, instead of using electromagnetic fields as most microwave engineering books do, the subject is introduced via circuit concepts. Further, an overview of communication systems is presented in the beginning to provide the reader with an overall perspective of various building blocks involved. ix

x

PREFACE

This edition of the book is organized into thirteen chapters and nine appendixes, using a top-down approach. It begins with an introduction to frequency bands, RF and microwave devices, and applications in communication, radar, industrial, and biomedical areas. The introduction includes a brief description of microwave transmission lines: waveguides, strip lines, and microstrip line. An overview of transmitters and receivers is included, along with digital modulation and demodulation techniques. Modern wireless communication systems, such as terrestrial and satellite communication systems and RF wireless services, are discussed briefly in Chapter 2. After introducing antenna terminology, effective isotropic radiated power, the Friis transmission formula, and the radar range equation are presented. In the final section of the chapter, noise and distortion associated with communication systems are introduced. Chapter 3 begins with a discussion of distributed circuits and construction of a solution to the transmission line equation. Topics presented in this chapter include RF circuit analysis, phase and group velocities, sending-end impedance, reflection coefficient, return loss, insertion loss, experimental determination of characteristic impedance and the propagation constant, the voltage standing wave ratio, and impedance measurement. The final section in Chapter 3 includes a description of the Smith chart and its application in the analysis of transmission line circuits. Fundamental laws of electromagnetic fields are introduced in Chapter 4 along with wave equations and uniform plane wave solutions. Boundary conditions and potential functions that lead to the construction of solutions are then introduced. The chapter concludes with analyses of various metallic waveguides. Resonant circuits are discussed in Chapter 5, which begins with series and parallel resistance–inductance–capacitance circuits. This is followed by a section on transformer-coupled circuits. The final two sections of the chapter are devoted to transmission line resonant circuits and microwave resonators. Chapters 6 and 7 deal with impedance-matching techniques. Single reactive element or stub, double-stub, and lumped-element matching techniques are discussed in Chapter 6. Chapter 7 is devoted to multisection transmission line impedance transformers, binomial and Chebyshev sections, and impedance tapers. Chapter 8 introduces circuit parameters associated with two-port networks. Impedance, admittance, hybrid, transmission, scattering, and chain scattering parameters are presented, along with examples that illustrate their characteristic behaviors. Chapter 9 begins with the image parameter method for the design of passive filter circuits. The insertion-loss technique is introduced next to synthesize Butterworth and Chebyshev low-pass filters. The chapter includes descriptions of impedance and frequency scaling techniques to realize high-pass, bandpass, and bandstop networks. The chapter concludes with a section on microwave transmission line filter design. Concepts of signal-flow-graph analysis are introduced in Chapter 10, along with a representation of voltage source and passive devices, which facilitates formulation of the power gain relations that are needed in the amplifier design discussed in Chapter 11. The chapter begins with stability considerations using

PREFACE

xi

scattering parameters of a two-port network followed by design techniques of various amplifiers. Chapter 12 presents basic concepts and design of various oscillator circuits. The phase-locked loop and its application in the design of frequency synthesizers are also summarized. The final section of the chapter includes the analysis and design of microwave transistor oscillators using S-parameters. Chapter 13 includes the fundamentals of frequency-division multiplexing, amplitude modulation, radio-frequency detection, frequency-modulated signals, and mixer circuits. The book ends with nine appendixes, which include a discussion of logarithmic units (dB, dBm, dBW, dBc, and neper), design equations for selected transmission lines (coaxial line, strip line, and microstrip line), and a list of abbreviations used in the communication area. Some of the highlights of the book are as follows:

ž

The presentation begins with an overview of frequency bands, RF and microwave devices, and their applications in various areas. Communication systems are presented in Chapter 2, including terrestrial and satellite systems, wireless services, antenna terminology, the Friis transmission formula, the radar equation, and Doppler radar. Thus, students learn about the systems using blocks of amplifiers, oscillators, mixers, filters, and so on. Students’ response has strongly supported this top-down approach.

ž

Since students are assumed to have only one semester of electrical circuits, resonant circuits and two-port networks are included in the book. Concepts of network parameters (impedance, admittance, hybrid, transmission, and scattering) and their characteristics are introduced via examples.

ž

A separate chapter on oscillator design includes concepts of feedback, the Hartley oscillator, the Colpitts oscillator, the Clapp oscillator, crystal oscillators, phased-locked-loop and frequency synthesizers, transistor oscillator design using S-parameters, and three-port S-parameter description of transistors and their use in feedback network design.

ž

A separate chapter on detectors and mixers includes amplitude- and frequency-modulated signal characteristics and their detection schemes, single-diode mixers, RF detectors, double-balanced mixers, conversion loss, intermodulation distortion in diode-ring mixers, and field-effecttransistor mixers.

ž

Appendixes include logarithmic units, design equations for selected transmission lines, and a list of abbreviations used in the communication area.

ž

There are 153 solved examples with a step-by-step explanation. Therefore, practicing engineers will find the book useful for self-study as well.

ž

There are 275 class-tested problems at the ends of chapters. Supplementary material is available to instructors adopting the book.

xii

PREFACE

Acknowledgments I learned this subject from engineers and authors who are too many to include in this short space, but I gratefully acknowledge their contributions. I would like to thank my anonymous reviewers, instructors here and abroad who used the first edition of this book and provided a number of constructive suggestions, and my former students, who made useful suggestions to improve the presentation. I deeply appreciate the support I received from my wife, Ila, and son, Shashank, during the course of this project. The first edition of this book became a reality only because of enthusiastic support from then-senior editor Philip Meyler and his staff at Wiley. I feel fortunate to continue getting the same kind of support from current editor Val Moliere and from Kirsten Rohstedt. DEVENDRA K. MISRA

1 INTRODUCTION

Scientists and mathematicians of the nineteenth century laid the foundation of telecommunication and wireless technology, which has affected all facets of modern society. In 1864, James C. Maxwell put forth fundamental relations of electromagnetic fields that not only summed up the research findings of Laplace, Poisson, Faraday, Gauss, and others but also predicted the propagation of electrical signals through space. Heinrich Hertz subsequently verified this in 1887, and Guglielmo Marconi transmitted wireless signals across the Atlantic Ocean successfully in 1900. Interested readers may find an excellent discussion of the historical developments of radio frequencies (RFs) and microwaves in the IEEE Transactions on Microwave Theory and Technique (Vol. MTT-32, September 1984). Wireless communication systems require high-frequency signals for the efficient transmission of information. Several factors lead to this requirement. For example, an antenna radiates efficiently if its size is comparable to the signal wavelength. Since the signal frequency is inversely related to its wavelength, antennas operating at RFs and microwaves have higher radiation efficiencies. Further, their size is relatively small and hence convenient for mobile communication. Another factor that favors RFs and microwaves is that the transmission of broadband information signals requires a high-frequency carrier signal. In the case of a single audio channel, the information bandwidth is about 20 kHz. If amplitude modulation (AM) is used to superimpose this information on a carrier, it requires at least this much bandwidth on one side of the spectrum. Further, Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

1

2

INTRODUCTION

TABLE 1.1 Frequency Bands Used in Commercial Broadcasting

AM TV FM TV

Channels

Frequency Range

Wavelength Range

107 2–4 5–6 100 7–13 14–83

535–1605 kHz 54–72 MHz 76–88 MHz 88–108 MHz 174–216 MHz 470–890 MHz

186.92–560.75 m 4.17–5.56 m 3.41–3.95 m 2.78–3.41 m 1.39–1.72 m 33.7–63.83 cm

commercial AM transmission requires a separation of 10-kHz between the two transmitters. On the other hand, the required bandwidth increases significantly if frequency modulation (FM) is used. Each FM transmitter typically needs a bandwidth of 200 kHz for audio transmission. Similarly, each television channel requires about 6 MHz of bandwidth to carry the video information as well. Table 1.1 shows the frequency bands used for commercial radio and television broadcasts. In the case of digital transmission, a standard monochrome television picture is sampled over a grid of 512 × 480 elements called pixels. Eight bits are required to represent 256 shades of the gray display. To display motion, 30 frames are sampled per second; thus, it requires about 59 Mb/s (512 × 480 × 8 × 30 = 58,982,400). Color transmission requires even higher bandwidth (on the order of 90 Mb/s). Wireless technology has been expanding very fast, with new applications reported every day. In addition to the traditional applications in communication, such as radio and television, RF and microwave signals are being used in cordless phones, cellular communication, local, wide, and metropolitan area networks and personal communication service. Keyless door entry, radio-frequency identification (RFID), monitoring of patients in a hospital or a nursing home, and cordless mice or keyboards for computers are some of the other areas where RF technology is being used. Although some of these applications have traditionally used infrared (IR) technology, current trends favor RF, because RF is superior to infrared technology in many ways. Unlike RF, infrared technology requires unobstructed line-of-sight connection. Although RF devices have been more expensive than IR, the current trend is downward because of an increase in their production and use. The electromagnetic frequency spectrum is divided into bands as shown in Table 1.2. Hence, AM radio transmission operates in the medium-frequency (MF) band, television channels 2 to 12 operate in the very high frequency (VHF) band, and channels 18 to 90 operate in the ultrahigh-frequency (UHF) band. Table 1.3 shows the band designations in the microwave frequency range. In addition to natural and human-made changes, electrical characteristics of the atmosphere affect the propagation of electrical signals. Figure 1.1 shows

3

INTRODUCTION

TABLE 1.2

IEEE Frequency Band Designations

Band Designation VLF LF MF HF VHF UHF SHF EHF

TABLE 1.3

Frequency Range 3–30 kHz 30–300 kHz 300–3000 kHz 3–30 MHz 30–300 MHz 300–3000 MHz 3–30 GHz 30–300 GHz

Wavelength Range (in Free Space) 10–100 km 1–10 km 100 m–1 km 10–100 m 1–10 m 10 cm–1 m 1–10 cm 0.1–1 cm

Microwave Frequency Band Designations

Frequency Bands 500–1000 MHz 1–2 GHz 2–4 GHz 3–4 GHz 4–6 GHz 6–8 GHz 8–10 GHz 10–12.4 GHz 12.4–18 GHz 18–20 GHz 20–26.5 GHz 26.5–40 GHz

Old (Still Widely Used)

New (Not So Commonly Used)

UHF L S S C C X X Ku K K Ka

C D E F G H I J J J K K

various layers of the ionosphere and the troposphere that are formed due to the ionization of atmospheric air. As illustrated in Figure 1.2(a) and (b), an RF signal can reach the receiver by propagating along the ground or after reflection from the ionosphere. These signals may be classified as ground and sky waves, respectively. The behavior of a sky wave depends on the season, day or night, and solar radiation. The ionosphere does not reflect microwaves, and the signals propagate line of sight, as shown in Figure 1.2(c). Hence, curvature of the Earth limits the range of a microwave communication link to less than 50 km. One way to increase the range is to place a human-made reflector up in the sky. This type of arrangement is called a satellite communication system. Another way to increase the range of a microwave link is to place repeaters at periodic intervals. This is known as a terrestrial communication system. Figures 1.3 and 1.4 list selected devices used at RF and microwave frequencies. Solid-state devices as well as vacuum tubes are used as active elements in RF and

4

INTRODUCTION F2 layer

F1 layer 300 km

E layer 200 km Ionosphere

125 km D layer 100 km 90 km 60 km

Troposphere 10 km

Figure 1.1

Atmosphere surrounding Earth.

microwave circuits. Predominant applications for microwave tubes are in radar, communications, electronic countermeasures (ECMs), and microwave cooking. They are also used in particle accelerators, plasma heating, material processing, and power transmission. Solid-state devices are employed primarily in the RF region and in low-power microwave circuits such as low-power transmitters for local area networks and receiver circuits. Some applications of solid-state devices are listed in Table-1.4. Figure 1.5 lists some applications of microwaves. In addition to terrestrial and satellite communications, microwaves are used in radar systems as well as in various industrial and medical applications. Civilian applications of radar include air traffic control, navigation, remote sensing, and law enforcement. Its military uses include surveillance; guidance of weapons; and command, control, and communication (C3 ). Radio-frequency and microwave energy are also used in industrial heating and for household cooking. Since this process does not use a conduction mechanism for the heat transfer, it can improve the quality of certain products significantly. For example, the hot air used in a printing press to dry ink affects the paper adversely and shortens the product’s life span. By contrast, in microwave drying only the ink portion is heated, and the paper is barely affected. Microwaves are also used in material processing, telemetry, imaging, and hyperthermia. 1.1 MICROWAVE TRANSMISSION LINES Figure 1.6 shows selected transmission lines used in RF and microwave circuits. The most common transmission line used in the RF and microwave range is the

5

MICROWAVE TRANSMISSION LINES

Receiving antenna

Transmitting antenna

Earth

(a) Signal propagation along the ground Ionosphere

Earth Receiving antenna

Transmitting antenna

(b) Propagation of a sky wave

Transmitting antenna

Earth

Receiving antenna

(c) Line-of-sight propagation

Figure 1.2

Modes of signal propagation.

coaxial line. A low-loss dielectric material is used in these transmission lines to minimize signal loss. Semirigid coaxial lines with continuous cylindrical conductors outside perform well in the microwave range. To ensure single-mode transmission, the cross section of a coaxial line must be much smaller than the signal wavelength. However, this limits the power capacity of these lines. In

6

INTRODUCTION Microwave Devices Active Solid-State

Vacuum-Tube Linear Beam Klystrons Traveling wave tubes Hybrid tubes

Passive Directional couplers Attenuators Cross-Field Phase shifters Magnetrons Antennas Resonators

Figure 1.3 Microwave devices. Solid-State Devices

Transistors

Transferred Electron

BJT FET HEMT

Gunn diode LSA diode InP diode CdTe diode

Figure 1.4 TABLE 1.4

Avalanche Transit Time BARITT diode IMPATT diode TRAPATT diode Parametric devices

Quantum Electronic Ruby masers Semiconductor laser

Solid-state devices used at RF and microwave frequencies.

Selected Applications of Microwave Solid-State Devices

Devices Transistors

Transferred electron devices (TED)

IMPATT diode

TRAPATT diode

BARITT diode

Applications

Advantages

L-band transmitters for telemetry systems and phased-array radar systems; transmitters for communication systems C-, X-, and Ku-band ECM amplifiers for wideband systems; X- and Ku-band transmitters for radar systems, such as traffic control Transmitters for millimeter-wave communication S-band pulsed transmitter for phased-array radar systems

Low cost, low power supply, reliable, high-continuouswave (CW) power output, lightweight

Local oscillators in communication and radar receivers

Low power supply (12 V), low cost, lightweight, reliable, low noise, high gain

Low power supply, low cost, reliable, high-CW power, lightweight High peak and average power, reliable, low power supply, low cost Low power supply, low cost, low noise, reliable

7

MICROWAVE TRANSMISSION LINES Microwave Applications Communication

Radar

Industrial and Biomedical

Terrestrial Civilian Air traffic Satellite control Aircraft navigation Ship safety

Military

Space vehicles Remote sensing

Heating

Surveillance

Process control Drying

Navigation

Curing

Guidance of weapons

Treatment of elastomers

Electronic warfare

Waste treatment

Industrial Household

Nuclear

C3

Cellulosic

Law enforcement

Monitoring Imaging Hyperthermia

Figure 1.5

Some applications of microwaves.

(b) Rectangular waveguide

(a) Coaxial line

(d) Ridged waveguide

(e) Fin line

εr (f) Strip line

Figure 1.6

(c) Circular waveguide

εr (g) Microstrip line

εr (h) Slot line

Transmission lines used in RF and microwave circuits.

high-power microwave circuits, waveguides are used in place of coaxial lines. Rectangular waveguides are commonly employed for connecting high-power microwave devices because these are easy to manufacture compared with circular waveguides. However, certain devices (e.g., rotary joints) require a circular cross section. In comparison with a rectangular waveguide, a ridged waveguide provides broadband operation. The fin line shown in Figure 1.6(e) is commonly

8

INTRODUCTION

used in the millimeter-wave band. Physically, it resembles a slot line enclosed in a rectangular waveguide. The transmission lines illustrated in Figure 1.6(f) to (h) are most convenient in connecting the circuit components on a printed circuit board (PCB). The physical dimensions of these transmission lines depend on the dielectric constant εr of insulating material and on the operating frequency band. The characteristics and design formulas of selected transmission lines are given in the appendixes.

1.2 TRANSMITTER AND RECEIVER ARCHITECTURES Wireless communication systems require a transmitter at one end to send the information signal and a receiver at the other to retrieve it. In one-way communication (such as a commercial broadcast), a transmitting antenna radiates the signal according to its radiation pattern. The receiver, located at the other end, receives this signal via its antenna and extracts the information, as illustrated in Figure 1.7. Thus, the transmitting station does not require a receiver, and vice versa. On the other hand, a transceiver (a transmitter and a receiver) is needed at both ends to establish a two-way communication link. Figure 1.7 is a simplified block diagram of a one-way communication link. At the transmitting end, an information signal is modulated and mixed with a local oscillator to up-convert the carrier frequency. Bandpass filters are used before

Up-converter Information signal

Modulator

BPF

BPF Power amplifier LO (a) Downconverter

Information signal

Demodulator

BPF/IF amp.

BPF

LO

Low-noise Frequencyamplifier selective circuit

(b)

Figure 1.7 Simplified block diagram of the transmitter (a) and receiver (b) of a wireless communication system.

TRANSMITTER AND RECEIVER ARCHITECTURES

9

and after the mixer to stop undesired harmonics. The signal power is amplified before feeding it to the antenna. At the receiving end the entire process is reversed to recover the information. Signal received by the antenna is filtered and amplified to improve the signal-to-noise ratio before feeding it to the mixer for down-converting the frequency. A frequency-selective amplifier (tuned amplifier) amplifies it further, before feeding it to a suitable demodulator, which extracts the information signal. In an analog communication system, the amplitude or angle (frequency or phase) of the carrier signal is varied according to the information signal. These modulations are known as amplitude modulation (AM), frequency modulation (FM), and phase modulation (PM), respectively. In a digital communication system, the input passes through channel coding, interleaving, and other processing before it is fed to the modulator. Various modulation schemes are available, including on–off keying (OOK), frequency-shift keying (FSK), and phase-shift keying (PSK). As these names suggest, a high-frequency signal is turned on and off in OOK to represent the logic states 1 and 0 of a digital signal. Similarly, two different signal frequencies are employed in FSK to represent the two signal states. In PSK, the phase of the high-frequency carrier is changed according to the 1 or 0 state of the signal. If only two phase states (0◦ and 180◦ ) of the carrier are used, it is called binary phase-shift keying (BPSK). On the other hand, a phase shift of 90◦ gives four possible states, each representing 2 bits of information (known as a dibit). This type of digital modulation, known as quadrature phase-shift keying (QPSK), is explained further below. Consider a sinusoidal signal Smod , as given by equation (1.2.1). Its angular frequency and phase are ω radians per second and φ, respectively: Smod (t) =

√

2 cos(ωt − φ) =

√

2(cos ωt cos φ + sin ωt sin φ)

(1.2.1)

This can be simplified further as follows: Smod (t) = SI cos ωt + SQ sin ωt where SI = and SQ =

√

2 cos φ

√

2 sin φ

(1.2.2)

(1.2.3)

(1.2.4)

The subscripts I and Q represent in-phase and quadrature-phase components of Smod . Table 1.5 shows the values of SI and SQ for four different phase states together with the corresponding dibit representations. This scheme can be implemented easily for a polar digital signal (positive peak value representing logic 1 and the negative peak logic 0). It is illustrated in Figure 1.8. The demultiplexer simultaneously feeds one bit of the digital input SBB to the top and the other to the bottom branch of this circuit. The top branch multiplies the signal by cos ωt

10

INTRODUCTION

TABLE 1.5 φ

QPSK Scheme

SI

SQ

π/4 1 3π/4 −1 5π/4 −1 7π/4 1

1 1 −1 −1

Dibit Representation 11 01 00 10

Logic 1 = 1 V Logic 0 = −1 V

A cos ωt

Demultiplexer SBB(t)

I

Q

LO

Σ

Smod(t)

90° Phase shift A sin ωt Logic 1 = 1 V Logic 0 = −1 V

Figure 1.8

Block diagram of a QPSK modulation scheme.

Threshold detector

SBB(t)

Integrator A cos ωt

Multiplexer I

Smod(t)

LO

Q

90° Phase shift Asin ωt Threshold detector

Integrator

Figure 1.9 Block diagram of a QPSK demodulation scheme.

and the bottom signal by sin ωt. The outputs of the two mixers are then added to generate Smod . Demodulation inverts the modulation to retrieve SBB . As illustrated in Figure 1.9, Smod is multiplied by cos ωt in the top branch and by sin ωt in the bottom branch of the demodulator. The top integrator stops SQ , and the bottom integrator, SI . Two threshold detectors generate the corresponding logic states that are multiplexed by the multiplexer unit to recover SBB . Chapter 2 provides an overview of wireless communication systems and their characteristics.

2 COMMUNICATION SYSTEMS

Modern communication systems require RF and microwave signals for the wireless transmission of information. These systems employ oscillators, mixers, filters, and amplifiers to generate and process various types of signals. The transmitter communicates with the receiver via antennas placed on each side. Electrical noise associated with the systems and the channel affects the performance. A system designer needs to know about the channel characteristics and system noise in order to estimate the required power levels. The chapter begins with an overview of microwave communication systems and RF wireless services to illustrate the applications of circuits and devices that are described in the following chapters. It also covers the placement of various building blocks in a given system. A short discussion on antennas is included to help in understanding signal behavior when it propagates from a transmitter to a receiver. The Friis transmission formula and the radar range equation are important to understanding the effects of frequency, range, and operating power levels on the performance of a communication system. Note that radar concepts now find many other applications, such as proximity or level sensing in an industrial environment. Therefore, a brief discussion of Doppler radar is also included. Noise and distortion characteristics play a significant role in analysis and design of these systems. Minimum detectable signal (MDS), gain compression, intercept point, and the dynamic range of an amplifier (or receiver) are introduced later. Other concepts associated with noise and distortion characteristics are also introduced in this chapter. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

11

12

COMMUNICATION SYSTEMS

2.1 TERRESTRIAL COMMUNICATION As mentioned in Chapter 1, microwave signals propagate along the line of sight. Therefore, the Earth’s curvature limits the range over which a microwave communication link can be established. A transmitting antenna sitting on a 25-ft-high tower can typically communicate only up to a distance of about 50 km. Repeaters can be placed at regular intervals to extend the range. Figure 2.1 is a block diagram of a typical repeater. A repeater system operates as follows. A microwave signal arriving at antenna A works as input to port 1 of a circulator. It is directed to port 2 without loss, assuming that the circulator is ideal. Then it passes through a receiver protection circuit that limits the magnitude of large signals but passes those of low intensity with negligible attenuation. The purpose of this circuit is to block excessively large signals from reaching the receiver input. The mixer following it works as a down-converter that transforms a high-frequency signal to a low-frequency signal, typically in the range of 70 MHz. A Schottky diode is generally employed in the mixer because of its superior noise characteristics. This frequency conversion facilitates amplification of the signal economically. A bandpass filter is used at the output of the mixer to stop undesired harmonics. An intermediate-frequency (IF) amplifier is then used to amplify the signal. It is generally a low-noise solid-state

Circulator B

Circulator A A

1

Transmitter for direction A

3

Received from direction B

1

2

3

2 Receiver protection circuit

Transreceiver for the reverse direction (from B to A)

Power amplifier

Mixer Mixer

Bandpass filter

IF amplifier with AGC

Limiter circuit

Bandpass filter

Mixer Bandpass filter

Power divider

Shift oscillator

Figure 2.1

Microwave source

Block arrangement of a repeater system.

B

SATELLITE COMMUNICATION

13

amplifier with ultralinear characteristics over a broadband. The amplified signal is mixed with another signal for up-conversion of frequency. After filtering out undesired harmonics introduced by the mixer, it is fed to a power amplifier stage that feeds circulator B for onward transmission through antenna B. This upconverting mixer circuit generally employs a varactor diode. Circulator B directs the signal entering at port 3 to the antenna connected at its port 1. Similarly, the signal propagating upstream is received by antenna B and the circulator directs it toward port 2. It then goes through the processing as described for the downstream signal and is radiated by antenna A for onward transmission. Hence, the downstream signal is received by antenna A and transmitted in the forward direction by antenna B. Similarly, the upstream signal is received by antenna B and forwarded to the next station by antenna A. The two circulators help channel the signal in the correct direction. A parabolic antenna with tapered horn as primary feeder is generally used in microwave links. This type of composite antenna system, known as a hog horn, is fairly common in high-density links because of its broadband characteristics. These microwave links operate in the frequency range 4 to 6 GHz, and signals propagating in two directions are separated by a few hundred megahertz. Since this frequency range overlaps with C-band satellite communication, the interference of these signals needs to be taken into design consideration. A single frequency can be used twice for transmission of information using vertical and horizontal polarization.

2.2 SATELLITE COMMUNICATION The ionosphere does not reflect microwaves as it does RF signals. However, one can place a conducting object (satellite) up in the sky that reflects them back to Earth. A satellite can even improve the signal quality using on-board electronics before transmitting it back. The gravitational force needs to be balanced somehow if this object is to stay in position. An orbital motion provides this balancing force. If a satellite is placed at low altitude, greater orbital force will be needed to keep it in position. These low- and medium-altitude satellites are visible from a ground station only for short periods. On the other hand, satellites placed at an altitude of about 36,000 km over the equator, called geosynchronous or geostationary satellites are visible from their shadows at all times. C-band geosynchronous satellites use between 5725 and 7075 MHz for their uplinks. The corresponding downlinks are between 3400 and 5250 MHz. Table 2.1 lists the downlink center frequencies of a 24-channel transponder. Each channel has a total bandwidth of 40 MHz; 36 MHz of that carries the information, and the remaining 4 MHz is used as a guard band. It is accomplished with a 500-MHz bandwidth using different polarization for the overlapping frequencies. The uplink frequency plan may be found easily after adding 2225 MHz to these downlink frequencies. Figure 2.2 illustrates the simplified block diagram of a C-band satellite transponder. A 6-GHz signal received from the Earth station

14

COMMUNICATION SYSTEMS

TABLE 2.1

C-Band Downlink Transponder Frequencies

Horizontal Polarization Channel

Vertical Polarization

Center Frequency (MHz)

Channel

Center Frequency (MHz)

3720 3760 3800 3840 3880 3920 3960 4000 4040 4080 4120 4160

2 4 6 8 10 12 14 16 18 20 22 24

3740 3780 3820 3860 3900 3940 3980 4020 4060 4100 4140 4180

1 3 5 7 9 11 13 15 17 19 21 23

Mixer BPF

Uplink 6-GHz signal

LNA

BPF

TWT amp.

LO

Figure 2.2

Downlink 4-GHz signal

Simplified block diagram of a transponder.

is passed through a bandpass filter before amplifying it through a low-noise amplifier (LNA). It is then mixed with a local oscillator (LO) signal to bring down its frequency. A bandpass filter that is connected right after the mixer filters out the unwanted frequency components. This signal is then amplified by a traveling wave tube (TWT) amplifier and transmitted back to Earth. Another frequency band in which satellite communication has been growing continuously is the Ku-band. The geosynchronous Fixed Satellite Service (FSS) generally operates between 10.7 and 12.75 GHz (space to Earth) and 13.75 to 14.5 GHz (Earth to space). It offers the following advantages over the C-band: ž ž

The size of the antenna can be smaller (3 ft or even smaller, with higherpower satellites against 8 to 10 ft for C-band). Because of higher frequencies used in the up- and downlinks, there is no interference with C-band terrestrial systems.

Since higher-frequency signals attenuate faster while propagating through adverse weather (rain, fog, etc.), Ku-band satellites suffer from this major

SATELLITE COMMUNICATION

15

drawback. Signals with higher powers may be used to compensate for this loss. Generally, this power is on the order of 40 to 60 W. The high-power direct broadcast satellite (DBS) system uses power amplifiers in the range 100 to 120 W. The National Broadcasting Company (NBC) has been using the Ku-band to distribute the programming to its affiliates. Also, various news-gathering agencies have used this frequency band for some time. Convenience stores, auto parts distributors, banks, and other businesses have used the very small aperture terminal (VSAT) because of its small antenna size (typically, on the order of 3 ft in diameter). It offers two-way satellite communication; usually back to hub or headquarters. The Public Broadcasting Service (PBS) uses VSATs for exchanging information among public schools. Direct broadcast satellites (DBSs) have been around since 1980, but early DBS ventures failed for various reasons. In 1991, Hughes Communications entered into the direct-to-home (DTH) television business. DirecTV was formed as a unit of GM Hughes, with DBS-1 launched in December 1993. Its longitudinal orbit is at 101.2◦ W, and it employs a left-handed circular polarization. DBS-2, launched in August 1994 uses a right-handed circular polarization, and its orbital longitude is at 100.8◦ W. DirecTV employs a digital architecture that can utilize video and audio compression techniques. It complies with Motion Picture Experts Group (MPEG)-2. By using compression ratios of 5 to 7, over 150 channels of programs are available from the two satellites. These satellites include 120-W TWT amplifiers that can be combined to form eight pairs at 240 W of power. This higher power can also be utilized for high-definition television (HDTV) transmission. Earth-to-satellite link frequency is 17.3 to 17.8 GHz; satellite-toEarth link frequency uses the 12.2- to 12.7-GHz band. Circular polarization is used because it is less affected by rain than is linear orthogonal polarization. Several communication services are now available that use low-Earth-orbit satellites (LEOSs) and medium-Earth-orbit satellites (MEOSs). LEOS altitudes range from 750 to 1500 km; MEOS systems have an altitude around 10,350 km. These services compete with or supplement cellular systems and geosynchronous Earth-orbit satellites (GEOSs). The GEOS systems have some drawbacks, due to the large distances involved. They require relatively large powers, and the propagation time delay creates problems in voice and data transmissions. The LEOS and MEOS systems orbit Earth faster because of being at lower altitudes, and these are therefore visible only for short periods. As Table 2.2 indicates, several satellites are used in a personal communication system to solve this problem. Three classes of service can be identified for mobile satellite services: 1. Data transmission and messaging from very small, inexpensive satellites 2. Voice and data communications from big LEOSs 3. Wideband data transmission Another application of L-band microwave frequencies (1227.60 and 1575.42 MHz) is in global positioning systems (GPSs). A constellation of 24 satellites is used to determine a user’s geographical location. Two services

16

COMMUNICATION SYSTEMS

TABLE 2.2

Specifications of Certain Personal Communication Satellites Iridium (LEO)

Number of satellites Altitude (km) Uplink (GHz) Downlink (GHz) Gateway terminal uplink (GHz) Gateway terminal downlink (GHz) Average satellite connect time (min) Features of handset Modulation BER Supportable data rate (kb/s)

Globalstar (LEO)

Odyssey (MEO)

66 755 1.616–1.6265 1.616–1.6265 27.5–30.0

48 1,390 1.610–1.6265 2.4835–2.500 C-band

12 10,370 1.610–1.6265 2.4835–2.500 29.5–30.0

18.8–20.2

C-band

19.7–20.2

9

10–12

120

QPSK 1E-2 (voice) 1E-5 (data) 4.8 (voice) 2.4 (data)

FQPSK 1E-3 (voice) 1E-5 (data) 1.2–9.6 (voice and data)

QPSK 1E-3 (voice) 1E-5 (data) 4.8 (voice) 1.2–9.6 (data)

are available: the standard positioning service (SPS) for civilian use, utilizing a single-frequency course/acquisition (C/A) code, and the precise positioning service (PPS) for the military, utilizing a dual-frequency P-code (protected). These satellites are at an altitude of 10,900 miles above the Earth, with an orbital period of 12 hours.

2.3 RADIO-FREQUENCY WIRELESS SERVICES A lot of exciting wireless applications have been reported that use voice and data communication technologies. Wireless communication networks consist of microcells that connect people with truly global, pocket-size communication devices, telephones, pagers, personal digital assistants, and modems. Typically, a cellular system employs a 100-W transmitter to cover a cell 0.5 to 10 miles in radius. The handheld transmitter has a power of less than 3 W. Personal communication networks (PCN/PCS) operate with a 0.01- to 1-W transmitter to cover a cell radius of less than 450 yards. The handheld transmitter power is typically less than 10 mW. Table 2.3 shows the cellular telephone standards of selected systems. There have been no universal standards set for wireless personal communication. In North America, cordless has been CT-0 (an analog 46/49-MHz standard) and cellular AMPS (Advanced Mobile Phone Service) operating at 800 MHz. The situation in Europe has been far more complex; every country has had its own standard. Although cordless was nominally CT-0, different countries used their own frequency plans. This led to a plethora of new standards. These include,

17

RADIO-FREQUENCY WIRELESS SERVICES

TABLE 2.3

Selected Cellular Telephones Analog Cellular Phone Standard NADC (IS-54)

NADC (IS-95)

GSM

PDC

824–849 871–904

824–849

824–849

880–915

869–894 916–949

869–894

869–894

925–960

940–956 1477–1501 810–826 1429–1453

600 mW

200 mW

1W

AMP Freq. range (MHz) Tx Rx Transmitter power (max.) Multiple access Number of channels Channel spacing (kHz) Modulation Bit rate (kb/s)

TABLE 2.4

Digital Cellular Phone Standard

ETACS

FDMA 832 30

FDMA TDMA/FDM CDMA/FDM TDMA/FDM TDMA/FDM 1000 832 20 124 1600 25 30 1250 200 25

FM —

FM —

π/4 DQPSK QPSK/BPSK 48.6 1228.8

π/4 DQPSK 42

GMSK 270.833

Selected Cordless Telephones Analog Cordless Phone Standard CT-0

CT-1 and CT-1+

Frequency range (MHz)

46/49

CT-1: 914/960; CT-1+: 885–932

Transmitter power (max.) (mW) Multiple access Number of channels Channel spacing (kHz) Modulation Bit rate (kb/s)

—

—

FDMA FDMA 10–20 CT-1: 40; CT-1+: 80 40 25 FM

FM

—

—

Digital Cordless Phone Standard CT-2 and CT-2+ CT-1: 864–868; CT-2+: 930/931; 940/941 10 and 80

TDMA/FDM 40

DECT

PHS (Formerly PHP)

1880–1900

1895–1918

250

80

100

TDMA/FDM TDMA/FDM 10 (12 users 300 (four users per channel) per channels) 1728 300

GFSK 72

GFSK 1152

π/4 DQPSK 384

but are not limited to, CT-1, CT-1+, DECT (Digital European Cordless Telephone), PHP (Personal Handy Phone in Japan), E-TACS (Extended Total Access Communication System in the UK), NADC (North American Digital Cellular), GSM (Global System for Mobile Communication), and PDC (Personal Digital Cellular). Specifications for selected cordless telephones are given in Table 2.4.

18

COMMUNICATION SYSTEMS

2.4 ANTENNA SYSTEMS Figure 2.3 illustrates some of the antennas that are used in communication systems. These can be categorized into two groups: wire antennas and aperture-type antennas. Electric dipole, monopole, and the loop antennas belong to the former group; horn, reflector, and lens belong to the latter category. The aperture antennas can be further subdivided into primary and secondary (or passive) antennas. Primary antennas are directly excited by the source and can be used independently for transmission or reception of signals. On the other hand, a secondary antenna requires another antenna as its feeder. Horn antennas fall in the first category, whereas the reflector and lens belong to the second. Various types of horn antennas are commonly used as feeders in reflector and lens antennas.

Electronics

Electronics (a)

(b)

(c)

Transmitter

(d )

(e)

Transmitter

(f)

Figure 2.3 Some commonly used antennas: (a) electric dipole; (b) monopole; (c) loop; (d) pyramidal horn; (e) Cassegrain reflector; (f ) lens.

ANTENNA SYSTEMS

19

When an antenna is energized, it generates two types of electromagnetic fields. Part of the energy stays nearby and part propagates outward. The propagating signal represents the radiation fields, while the nonpropagating is reactive (capacitive or inductive) in nature. Space surrounding the antenna can be divided into three regions. The reactive fields dominate in the nearby region but are reduced in strength at a faster rate than those associated with the propagating signal. If the largest dimension of an antenna is D and the signal wavelength is λ, reactive fields dominate up to about 0.62 D 3 /λ and diminish after 2D 2 /λ. The region beyond 2D 2 /λ is called the far-field (or radiation field) region. Power radiated by an antenna per unit solid angle is known as the radiation intensity U . It is a far-field parameter that is related to power density (power per unit area) Wrad and distance r as follows: U = r 2 Wrad

(2.4.1)

Directive Gain and Directivity If an antenna radiates uniformly in all directions, it is called an isotropic antenna. This is a hypothetical antenna that helps in defining the characteristics of a real one. The directive gain DG is defined as the ratio of radiation intensity due to the test antenna to that of an isotropic antenna. It is assumed that total radiated power remains the same in the two cases. Hence, DG =

U 4πU = U0 Prad

(2.4.2)

where U is the radiation intensity due to the test antenna in watts per unit solid angle, U0 the radiation intensity due to the isotropic antenna in watts per unit solid angle, and Prad the total power radiated in watts. Since U is a directionaldependent quantity, the directive gain of an antenna depends on the angles θ and φ. If the radiation intensity assumes its maximum value, the directive gain is called the directivity D0 . That is, D0 =

Umax 4πUmax = U0 Prad

(2.4.3)

Gain of an Antenna The power gain of an antenna is defined as the ratio of its radiation intensity at a point to the radiation intensity that results from a uniform radiation of the same input power. Hence, gain = 4π

radiation intensity U (θ, φ) = 4π total input power Pin

(2.4.4)

Most of the time we deal with relative gain. It is defined as a ratio of the power gain of the test antenna in a given direction to the power gain of a reference

20

COMMUNICATION SYSTEMS

antenna. Both antennas must have the same input power. The reference antenna is usually a dipole, horn, or any other antenna whose gain can be calculated or is known. However, the reference antenna is a lossless isotropic radiator in most cases. Hence, U (θ, φ) gain = 4π (2.4.5) Pin (lossless isotropic antenna) When the direction is not stated, the power gain is usually taken in the direction of maximum radiation. Radiation Patterns and Half-Power Beam Width Far-field power distribution at a distance r from the antenna depends on the spatial coordinates θ and φ. Graphical representations of these distributions on the orthogonal plane (θ-plane or φ-plane) at a constant distance r from the antenna are called its radiation patterns. Figure 2.4 illustrates the radiation pattern of the vertical dipole antenna with θ. Its φ-plane pattern can be found after rotating it about the vertical axis. Thus, a three-dimensional picture of the radiation pattern of a dipole is doughnut shaped. Similarly, the power distributions of other antennas generally show peaks and valleys in the radiation zone. The highest peak between the two valleys is known as the main lobe; the others are called side lobes. The total angle about the main peak over which power is reduced by 50% of its maximum value is called the half-power beam width on that plane.

θ

r 0

1

Figure 2.4 Radiation pattern of a dipole in a vertical (θ) plane.

ANTENNA SYSTEMS

21

The following relations are used to estimate the power gain G and the halfpower beam width (HPBW, or simply BW) of an aperture antenna: G=

4π 4π Ae = 2 Aκ 2 λ λ

(2.4.6)

65λ d

(2.4.7)

and BW(degrees) =

where Ae is the effective area of radiating aperture in square meters, A its physical area (πd 2 /4 for a reflector antenna dish with diameter d), κ the antenna efficiency (ranges from 0.6 to 0.65), and λ the signal wavelength in meters. Example 2.1 Calculate the power gain (in decibels) and the half-power beam width of a parabolic dish antenna 30 m in diameter that is radiating at 4 GHz. SOLUTION The signal wavelength and area of the aperture are λ=

3 × 108 = 0.075 m 4 × 109

A=

πd 2 302 =π = 706.8584 m2 4 4

and

Assuming that the aperture efficiency is 0.6, the antenna gain and half-power beam width are found as follows: G=

4π × 706.8584 × 0.6 = 947,482.09 (0.075)2 = 10 log10 (947,482.09) = 59.76 ≈ 60 dB

BW =

65 × 0.075 ◦ = 0.1625 30

Antenna Efficiency If an antenna is not matched with its feeder, a part of the signal available from the source is reflected back. It is considered to be the reflection (or mismatch) loss. The reflection (or mismatch) efficiency is defined as a ratio of power input to the antenna to that of power available from the source. Since the ratio of reflected power to that of power available from the source is equal to the square of the magnitude of the voltage reflection coefficient, the reflection efficiency er is given by er = 1 − ||2

22

COMMUNICATION SYSTEMS

where = voltage reflection coefficient =

ZA − Z0 ZA + Z0

Here ZA is the antenna impedance and Z0 is the characteristic impedance of the feeding line. In addition to mismatch, the signal energy may dissipate in an antenna due to imperfect conductor or dielectric material. These efficiencies are hard to compute. However, the combined conductor and dielectric efficiency ecd can be determined experimentally after measuring the input power Pin and the radiated power Prad . It is given as Prad ecd = Pin The overall efficiency eo is a product of the efficiencies above. That is, eo = er ecd

(2.4.8)

Example 2.2 A 50- transmission line feeds a lossless one-half-wavelengthlong dipole antenna. The antenna impedance is 73 . If its radiation intensity, U (θ, φ), is given as U = B0 sin3 θ find the maximum overall gain. SOLUTION The maximum radiation intensity, Umax , is the B0 value that occurs at θ = π/2. Its total radiated power is found as follows:

2π

Prad = 0

π 0

B0 sin3 θ sin θ dθ dφ = 34 π2 B0

Hence, D0 = 4π

Umax 4πB0 16 = 1.6977 = 3 2 = Prad 3π π B0 4

or D0 (dB) = 10 log10 (1.6977)dB = 2.2985 dB Since the antenna is lossless, the radiation efficiency ecd is unity (0 dB). Its mismatch efficiency is computed as follows. The voltage reflection coefficient at its input (it is formulated in Chapter 2) is =

23 ZA − Z0 73 − 50 = = ZA + Z0 73 + 50 123

ANTENNA SYSTEMS

23

Therefore, the mismatch efficiency of the antenna is er = 1 − (23/123)2 = 0.9650 = 10 log10 (0.9650)dB = −0.1546 dB The overall gain G0 (in decibels) is found as follows: G0 (dB) = 2.2985 − 0 − 0.1546 = 2.1439 dB Bandwidth Antenna characteristics such as gain, radiation pattern, impedance, and so on are frequency dependent. The bandwidth of an antenna is defined as the frequency band over which its performance with respect to some characteristic (HPBW, directivity, etc.) conforms to a specified standard. Polarization Polarization of an antenna is the same as polarization of its radiating wave. It is a property of the electromagnetic wave describing the time-varying direction and relative magnitude of an electric field vector. The curve traced by the instantaneous electric field vector with time is the polarization of that wave. The polarization is classified as follows: ž ž

ž

Linear polarization. If the tip of the electric field intensity traces a straight line in some direction with time, the wave is linearly polarized. Circular polarization. If the end of the electric field traces a circle in space as time passes, that electromagnetic wave is circularly polarized. Further, it may be right-handed circularly polarized (RHCP) or left-handed circularly polarized (LHCP), depending on whether the electric field vector rotates clockwise or counterclockwise. Elliptical polarization. If the tip of the electric field intensity traces an ellipse in space as time lapses, the wave is elliptically polarized. As in the preceding case, it may be right- or left-handed elliptical polarization (RHEP and LHEP).

In a receiving system, the polarization of the antenna and the incoming wave need to be matched for maximum response. If this is not the case, there will be some signal loss, known as polarization loss. For example, if there is a vertically polarized wave incident on a horizontally polarized antenna, the induced voltage available across its terminals will be zero. In this case, the antenna is crosspolarized with an incident wave. The square of the cosine of the angle between wave polarization and antenna polarization is a measure of the polarization loss. It can be determined by squaring the scalar product of unit vectors representing the two polarizations.

24

COMMUNICATION SYSTEMS

Example 2.3 The electric field intensity of an electromagnetic wave propagating in a lossless medium in the z-direction is given by E(r, t) = xE ˆ 0 (x, y) cos(ωt − kz)

V/m

It is incident upon an antenna that is linearly polarized as follows: Ea (r) = (xˆ + y)E(x, ˆ y, z)

V/m

Find the polarization loss factor. SOLUTION In this case, the incident wave is linearly polarized along the x-axis while the receiving antenna is linearly polarized at 45◦ from it. Therefore, onehalf of the incident signal is cross-polarized with the antenna. It is determined mathematically as follows. The unit vector along the polarization of incident wave is uˆ i = xˆ The unit vector along the antenna polarization may be found as 1 uˆ a = √ (xˆ + y) ˆ 2 Hence, the polarization loss factor is |uˆ i •uˆ a |2 = 0.5 = −3.01 dB Effective Isotropic Radiated Power The effective isotropic radiated power (EIRP) is a measure of the power gain of the antenna. It is equal to the power needed by an isotropic antenna that provides the same radiation intensity at a given point as that of the directional antenna. If power input to the feeding line is Pt and the antenna gain is Gt , the EIRP is defined as Pt Gt EIRP = (2.4.9) L where L is the input/output power ratio of the transmission line that is connected between the output of the final power amplifier stage of the transmitter and the antenna. It is given by Pt L= (2.4.10) Pant Alternatively, the EIRP can be expressed in dBw as EIRP(dBw) = Pt (dBw) − L(dB) + G(dB)

(2.4.11)

ANTENNA SYSTEMS

25

Example 2.4 In a transmitting system, the output of the final high-power amplifier is 500 W, and the line feeding its antenna has an attenuation of 20%. If the gain of the transmitting antenna is 60 dB, find the EIRP in dBw. SOLUTION Pt = 500 W = 26.9897 dBw Pant = 0.8 × 500 = 400 W G = 60 dB = 106 and L=

500 = 1.25 = 10 log10 (1.25) = 0.9691 dB 400

Hence, EIRP(dBw) = 26.9897 − 0.9691 + 60 = 86.0206 dBw or EIRP =

500 × 106 = 400 × 106 W 1.25

Space Loss The transmitting antenna radiates in all directions, depending on its radiation characteristics. However, the receiving antenna receives only the power that is incident on it. Hence, the rest of the power is not used and is lost in space. It is represented by the space loss. It can be determined as follows. The power density wt of a signal transmitted by an isotropic antenna is given by Pt wt = W/m2 (2.4.12) 4πR 2 where Pt is the transmitted power in watts and R is the distance from the antenna in meters. The power received by a unity-gain antenna located at R is found to be Pr = wt Aeu where Aeu is the effective area of an isotropic antenna. From (2.4.6), for an isotropic antenna, G=

4π Aeu = 1 λ2

or Aeu =

λ2 4π

(2.4.13)

26

COMMUNICATION SYSTEMS

Hence, (2.4.12) can be written as Pr =

Pt λ2 4πR 2 4π

(2.4.14)

and the space loss ratio is found to be Pr = Pt

λ 4πR

2 (2.4.15)

It is usually expressed in decibels as follows: space loss ratio = 20 log10

λ 4πR

dB

(2.4.16)

Example 2.5 A geostationary satellite is 35,860 km away from Earth’s surface. Find the space loss ratio if it is operating at 4 GHz. SOLUTION R = 35,860,000 m

and λ =

3 × 108 = 0.075 m 4 × 109

Hence, space loss ratio =

4π × 35,860,000 0.075

2

= 2.77 × 10−20 = −195.5752 dB

Friis Transmission Formula and Radar Range Equation Analysis and design of communication and monitoring systems often require an estimation of transmitted and received powers. The Friis transmission formula and radar range equation provide the means for such calculations. The former is applicable to a one-way communication system where the signal is transmitted at one end and is received at the other end of the link. The radar range equation is applicable when the signal transmitted hits a target and the signal reflected is generally received at the location of the transmitter. We consider these two formulations next. Friis Transmission Equation Consider a simplified communication link as illustrated in Figure 2.5. A distance R separates the transmitter and the receiver. Effective apertures of transmitting and receiving antennas are Aet and Aer , respectively. Further, the two antennas

27

ANTENNA SYSTEMS

R

Transmitter

Figure 2.5

Receiver

Simplified block diagram of a communication link.

are assumed to be polarization matched. If power input to the transmitting antenna is Pt , isotropic power density w0 at a distance R from the antenna is given as w0 =

Pt et 4πR 2

(2.4.17)

where et is the radiation efficiency of the transmitting antenna. For a directional transmitting antenna, the power density wt can be written as wt =

Pt Gt Pt et Dt = 4πR 2 4πR 2

(2.4.18)

where Gt is the gain and Dt is the directivity of transmitting antenna. The power collected by the receiving antenna is Pr = Aer wt

(2.4.19)

λ2 Gr 4π

(2.4.20)

From (2.4.6), Aer =

where the receiving antenna gain is Gr . Therefore, we find that Pr = or Pr = Pt

λ2 λ2 Pt Gt Gr wt = Gr 4π 4π 4πR 2

λ 4πR

2 Gr Gt = er et

λ 4πR

2 Dr Dt

(2.4.21)

If signal frequency is f , for a free-space link, λ 3 × 108 = 4πR 4πf R where f is in hertz and R is in meters.

(2.4.22)

28

COMMUNICATION SYSTEMS

Generally, the link distance is long and the signal frequency is high, such that kilometer and megahertz will be more convenient units than the usual meter and hertz, respectively. For R in kilometers and f in megahertz, we find that λ 0.3 3 × 108 1 = = 6 3 4πR 4π × 10 fMHz × 10 Rkm 4π fMHz Rkm

(2.4.23)

Hence, from (2.4.21), Pr (dBm) = Pt (dBm) + 20 log10

0.3 − 20 log10 (fMHz Rkm ) + Gt (dB) + Gr (dB) 4π

or Pr (dBm) = Pt (dBm) + Gt (dB) + Gr (dB) − 20 log10 (fMHz Rkm ) − 32.4418 (2.4.24) where the power transmitted and the power received are in dBm while the two antenna gains are in decibels. Example 2.6 A 20-GHz transmitter on board the satellite uses a parabolic antenna that is 45.7 cm in diameter. The antenna gain is 37 dB and its radiated power is 2 W. The ground station that is 36,941.031 km away from it has an antenna gain of 45.8 dB. Find the power collected by the ground station. How much power would be collected at the ground station if there were isotropic antennas on both sides? SOLUTION The power transmitted, Pt (dBm) = 10 log10 (2000) = 33.0103 dBm and 20 log10 (fMHz Rkm ) = 20 log10 (20 × 103 × 36,941.031) = 177.3708 dB Hence, the power received at the Earth station is found as follows: Pr (dBm) = 33.0103 + 37 + 45.8 − 177.3708 − 32.4418 = −94.0023 dBm or Pr = 3.979 × 10−10 mW If the two antennas are isotropic, Gt = Gr = 1 (or 0 dB) and therefore Pr (dBm) = 33.0103 + 0 + 0 − 177.3708 − 32.4418 = −176.8023 dBm or Pr = 2.0882 × 10−18 mW

29

ANTENNA SYSTEMS

Radar Equation In the case of a radar system, the signal transmitted is scattered by the target in all possible directions. The receiving antenna collects part of the energy that is scattered back toward it. Generally, a single antenna is employed for both the transmitter and the receiver, as shown in Fig. 2.6. If power input to the transmitting antenna is Pt and its gain is Gt , the power density winc incident on the target is Pt Gt Pt Aet winc = = 2 2 (2.4.25) 2 4πR λ R where Aet is the effective aperture of the transmitting antenna. The radar cross section σ of an object is defined as the area intercepting that amount of power that when scattered isotropically produces at the receiver a power density that is equal to that scattered by the actual target. Hence, radar cross section =

scattered power incident power density

or σ=

4πr 2 wr winc

square meters

(2.4.26)

where wr is isotropically backscattered power density at a distance r and winc is power density incident on the object. Hence, the radar cross section of an object is its effective area that intercepts an incident power density winc and gives an isotropically scattered power of 4πr 2 wr for a backscattered power density. Radar cross sections of selected objects are listed in Table 2.5. Using the radar cross section of a target, the power intercepted by it can be found as follows: σPt Gt Pinc = σwinc = (2.4.27) 4πR 2 Power density arriving back at the receiver is wscatter =

Pinc 4πR 2

Transmitter

Circulator Receiver

Figure 2.6

Antenna

Radar system.

(2.4.28)

30

COMMUNICATION SYSTEMS

TABLE 2.5

Radar Cross Sections of Selected Objects Radar Cross Section (m2 )

Object Pickup truck Automobile Jumbo-jet airliner Large bomber Large fighter aircraft Small fighter aircraft Adult male Conventional winged missile Bird Insect Advanced tactical fighter

200 100 100 40 6 2 1 0.5 0.01 0.00001 0.000001

and power available at the receiver input is Pr = Aer wscatter =

Gr λ2 σPt Gt σAer Aet Pt = 2 2 4π(4πR ) 4πλ2 R 4

(2.4.29)

Example 2.7 A distance of 100λ separates two lossless X-band horn antennas (Figure 2.7). Reflection coefficients at the terminals of transmitting and receiving antennas are 0.1 and 0.2, respectively. Maximum directivities of the transmitting and receiving antennas are 16 and 20 dB, respectively. Assuming that the input power in a lossless transmission line connected to the transmitting antenna is 2 W and that the two antennas are aligned for maximum radiation between them and are polarization matched, find the power input to the receiver. SOLUTION As discussed in Chapter 3, impedance discontinuity generates an echo signal very similar to that of an acoustical echo. Hence, signal power available beyond the discontinuity is reduced. The ratio of the reflected signal voltage to that of the incident is called the reflection coefficient. Since the power is proportional to the square of the voltage, the power reflected from the discontinuity is equal to the square of the reflection coefficient times the incident power.

100l Receiver

Transmitter

Pd = [1 − |Γ|2]PR Pt = [1 − |Γ|2]PS

Figure 2.7

Setup for Example 2.7.

ANTENNA SYSTEMS

31

Therefore, power transmitted in the forward direction will be given by Pt = (1 − ||2 )Pin and the power radiated by the transmitting antenna is found to be Pt = (1 − 0.12 )2 = 1.98 W Since the Friis transmission equation requires the antenna gain as a ratio instead of in decibels, Gt and Gr are calculated as follows: Gt = 16 dB = 101.6 = 39.8107

and Gr = 20 dB = 102.0 = 100

Hence, from (2.4.21), Pr =

λ 4π × 100λ

2 × 100 × 39.8107 × 1.98

or Pr = 5 mW and power delivered to the receiver, Pd , is Pd = (1 − 0.22 )5 = 4.8 mW Example 2.8 Radar operating at 12 GHz transmits 25 kW through an antenna of 25 dB gain. A target with its radar cross section at 8 m2 is located at 10 km from the radar. If the same antenna is used for the receiver, determine the power received. SOLUTION Pt = 25 kW f = 12 GHz → λ =

3 × 108 = 0.025 m 12 × 109

Gr = Gt = 25 dB → 102.5 = 316.2278 R = 10 km

σ = 8 m2

Hence, Pr =

Gr Gt Pt σλ2 316.22782 × 25, 000 × 8 × 0.0252 = = 6.3 × 10−13 W 4π(4πR 2 )2 (4π)3 (104 )4

or Pr = 0.63 pW

32

COMMUNICATION SYSTEMS

Doppler Radar An electrical signal propagating in free space can be represented by a simple expression as follows: v(z, t) = A cos(ωt − kz) (2.4.30) The signal frequency is ω radians per second and k is its wave number (equal to ω/c, where c is the speed of light in free space) in radians per meter. Assume that there is a receiver located at z = R, as shown in Figure 2.5 and R is changing with time (the receiver may be moving toward or away from the transmitter). In this situation, the receiver response vo (t) is given as follows: vo (t) = V cos(ωt − kR)

(2.4.31)

The angular frequency, ω0 , of vo (t) can be determined easily after differentiating the argument of the cosine function with respect to time. Hence, ω0 =

d dR (ωt − kR) = ω − k dt dt

(2.4.32)

Note that k is time independent and that the time derivative of R represents the velocity, vr , of the receiver with respect to the transmitter. Hence, (2.4.32) can be written ωvr vr (2.4.33) ω0 = ω − =ω 1− c c If the receiver is closing in, vr will be negative (negative slope of R), and therefore the signal received will indicate a signal frequency higher than ω. On the other hand, it will show a lower frequency if R is increasing with time. It is the Doppler frequency shift that is employed to design the Doppler radar. Consider the simplified block diagram shown in Figure 2.8. A microwave signal generated by the oscillator is split into two parts via a power divider. The circulator feeds one part of this power to an antenna that illuminates a target while a mixer uses the remaining fraction as its reference signal. Further, the antenna intercepts a part of the signal that is scattered by an object. It is then directed to the mixer through a circulator. The mixer output includes a difference frequency signal that can be filtered out for further processing. Two inputs to the mixer will have the same frequency if the target is stationary, and therefore the Doppler shift δω will be zero. On the other hand, the mixer output will have Doppler frequency if the target is moving. Note that the signal travels twice over the same distance, and therefore the Doppler frequency shift in this case will be twice that found via (2.4.33). Mathematically,

2vr ω0 = ω 1 − c

(2.4.34)

33

NOISE AND DISTORTION Circulator Oscillator

ω

ω

Power divider

ω0

ω ω0

Mixer δω

Amplifier

Display

Figure 2.8 Simplified block diagram of a Doppler radar.

and δω =

2ωvr c

(2.4.35)

2.5 NOISE AND DISTORTION Random movement of charges or charge carriers in an electronic device generates currents and voltages that vary randomly with time. In other words, the amplitude of these electrical signals cannot be predicted at any time. However, it can be expressed in terms of probability density functions. These signals are termed noise. For most applications it suffices to know the mean-square or root-meansquare value. Since the square of the current or the voltage is proportional to the power, mean-square noise voltage and current values are generally called noise power. Further, noise power is normally a function of frequency and the power per unit frequency (watts per hertz) is defined as the power spectral density of noise. If the noise power is the same over the entire frequency band of interest, it is called white noise. There are several mechanisms that can cause noise in an electronic device. Some of these are as follows: ž

ž

Thermal noise. This is the most basic type of noise, which is caused by thermal vibration of bound charges. Johnson studied this phenomenon in 1928, and Nyquist formulated an expression for spectral density around the same time. Therefore, it is also known as Johnson noise or Nyquist noise. In most electronic circuits, thermal noise dominates; therefore, it will be described further because of its importance. Shot noise. This is due to random fluctuations of charge carriers that pass through the potential barrier in an electronic device. For example, electrons

34

COMMUNICATION SYSTEMS

emitted from the cathode of thermionic devices or charge carriers in Schottky diodes produce a current that fluctuates about the average value I . The mean-square current due to shot noise is generally given by 2 iSh = 2eIB

ž

(2.5.1)

where e is electronic charge (1.602 × 10−19 C) and B is the bandwidth in hertz. Flicker noise. This occurs in solid-state devices and vacuum tubes operating at low frequencies. Its magnitude decreases with an increase in frequency. It is generally attributed to chaos in the dynamics of a system. Since the flicker noise power varies inversely with frequency, it is often called 1/f noise. Sometimes it is referred to as pink noise.

Thermal Noise Consider a resistor R that is at a temperature of T Kelvin. Electrons in this resistor are in random motion with a kinetic energy that is proportional to the temperature T . These random motions produce small, random voltage fluctuations across its terminals. This voltage has a zero average value but a nonzero mean-square value vn2 . It is given by Planck’s blackbody radiation law as vn2 =

4hfRB exp(hf/kT ) − 1

(2.5.2)

where h is Planck’s constant (6.546 × 10−34 J · s), k the Boltzmann constant (1.38 × 10−23 J/K), T the temperature in kelvin, B the system bandwidth in hertz, and f the center frequency of the bandwidth in hertz. For frequencies below 100 GHz, the product hf will be smaller than 6.546 × 10−23 J and kT will be greater than 1.38 × 10−22 J if T stays above 10 K. Therefore, kT will be larger than hf for such cases. Hence, the exponential term in equation (2.5.2) can be approximated as follows:

hf exp kT

≈1+

hf kT

Therefore, vn2 ≈

4hf RB = 4BRkT hf/kT

(2.5.3)

This is known as the Rayleigh–Jeans approximation. A Th´evenin-equivalent circuit can replace the noisy resistor as shown in Figure 2.9. As illustrated, it consists of a noise-equivalent voltage source in series with a noise-free resistor. This source will supply a maximum power to a load

35

NOISE AND DISTORTION

R Noise-free resistor

R Noisy resistor

Figure 2.9 Noise-equivalent circuit of a resistor.

of resistance R. The power delivered to that load in a bandwidth B is found as follows: v 2 Pn = n = kT B (2.5.4) 4R Conversely, if an arbitrary white noise source with its driving point impedance R delivers a noise power Ps to a load R, it can be represented by a noisy resistor of value R that is at temperature Te . Hence, Te =

Ps kB

(2.5.5)

where Te is an equivalent temperature selected so that the same noise power is delivered to the load. Consider the noisy amplifier as shown in Figure 2.10. Its gain is G over the bandwidth B. Let the amplifier be matched to the noiseless source and the load resistors. If the source resistor is at a hypothetical temperature of Ts = 0 K, the power input to the amplifier Pi will be zero and the output noise power Po will only be due to noise generated by the amplifier. We can obtain the same noise power at the output of an ideal noiseless amplifier by raising the temperature Ts of the source resistor to Te : Po Te = (2.5.6) GkB Hence, the output power in both cases is Po = GkTe B. The temperature Te is known as the equivalent noise temperature of the amplifier.

Ts = 0 K

R

Noisy amplifier

R

Te =

Po GkB

R

Noiseless amplifier

Figure 2.10 Noise-equivalent representation of an amplifier.

R

36

COMMUNICATION SYSTEMS

Measurement of Noise Temperature by the Y -Factor Method According to the definition, the noise temperature of an amplifier (or any other two-port network) can be determined by setting the source resistance R at 0 K and then measuring the output noise power. However, a temperature of 0 K cannot be achieved in practice. We can circumvent this problem by repeating the experiment at two different temperatures. This procedure is known as the Y-factor method. Consider an amplifier with a power gain of G over a frequency band of B hertz. Further, its equivalent noise temperature is Te (K). The input port of the amplifier is terminated by a matched resistor R while a matched power meter is connected at its output, as illustrated in Figure 2.11. With R at temperature Th , the power meter measures the noise output as P1 . Similarly, the noise power is found to be P2 when the temperature of R is set at Tc . Hence, P1 = GkTh B + GkTe B and P2 = GkTc B + GkTe B For Th higher than Tc , the noise power P1 will be larger than P2 . Therefore, P1 Th + Te =Y = P2 Tc + Te or Te =

Th − Y Tc Y −1

(2.5.7)

For Th larger than Tc , Y will be greater than unity. Further, measurement accuracy is improved by selecting two temperature settings that are far apart. Therefore, Th and Tc represent hot and cold temperatures, respectively. Example 2.9 An amplifier has a power gain of 10 dB in the 500-MHz to 1.5GHz frequency band. The following data are obtained for this amplifier using the

Th Amplifier P1

R

G, B, Te Tc

P2

R

Figure 2.11 Experimental setup for measurement of noise temperature.

NOISE AND DISTORTION

37

Y -factor method: At Th = 290 K, P1 = −70 dBm; at Tc = 77 K, P2 = −75 dBm. Determine its equivalent noise temperature. If this amplifier is used with a source that has an equivalent noise temperature of 450 K, find the output noise power in dBm. SOLUTION Since P1 and P2 are given in dBm, the difference of these two values will give Y in decibels. Hence, Y = (P1 − P2 ) dBm = (−70) − (−75) = 5 dB or Y = 100.5 = 3.1623 Therefore, Te =

290 − (3.1623)(77) = 21.51 K 3.1623 − 1

If a source with an equivalent noise temperature of Ts = 450 K drives the amplifier, the noise power input to this will be kTs B. The total noise power at the output of the amplifier will be Po = GkTs B + GkTe B = 10 × 1.38 × 10−23 × 109 × (450 + 21.51) = 6.5068 × 10−11 W Therefore,

Po = 10 log(6.5068 × 10−8 ) = −71.8663 dBm

Noise Factor and Noise Figure The noise factor of a two-port network is obtained by dividing the signal-to-noise ratio at its input port by the signal-to-noise ratio at its output. Hence, noise factor, F =

Si /Ni So /No

where Si , Ni , So , and No represent the power in input signal, input noise, output signal, and output noise, respectively. If the two-port network is noise-free, the signal-to-noise ratio at its output will be the same as its input, resulting in a noise factor of unity. In reality, the network will add its own noise, while the input signal and noise will be altered by the same factor (gain or loss). It will lower the output signal/noise ratio, resulting in a higher noise factor. Therefore, the noise factor of a two-port network is generally greater than unity. By definition, the input noise power is assumed to be the noise power resulting from a matched resistor at T0 = 290 K (i.e., Ni = kT0 B). Using the circuit arrangement illustrated

38

COMMUNICATION SYSTEMS T0 = 290 K

R Noisy two-port network G, B, Te Pi = Si + Ni

R Po = So + No

Figure 2.12 Circuit arrangement for determination of the noise factor of a noisy two-port network.

in Figure 2.12, the noise factor of a noisy two-port network can be defined as follows: noise factor =

total output noise in B when input source temperature is 290 K output noise of source (only) at 290 K

or F =

Pint kT0 BG + Pint =1+ kT0 BG kT0 BG

(2.5.8)

where Pint represents the output noise power that is generated by the two-port network internally. It can be expressed in terms of noise factor as follows: Pint = k(F − 1)T0 BG = kTe BG

(2.5.9)

where Te is known as the equivalent noise temperature of a two-port network. It is related to the noise factor as follows: Te = (F − 1)T0

(2.5.10)

When the noise factor is expressed in decibels, it is commonly called the noise figure (NF). Hence, NF = 10 log10 F dB (2.5.11) Example 2.10 The power gain of an amplifier is 20 dB in the frequency band 10 to 12 GHz. If its noise figure is 3.5 dB, find the output noise power in dBm. SOLUTION Output noise power = kT0 BG + Pint = F kT0 BG = No F = 3.5 dB = 100.35 = 2.2387 G = 20 dB = 102 = 100 Therefore, No = 2.2387 × 1.38 × 10−23 × 290 × 2 × 109 × 100 = 1.7919 × 10−9 W

39

NOISE AND DISTORTION

or No = 10 log10 (1.7919 × 10−6 mW) dBm = −57.4669 dBm ≈ −57.5 dBm Noise in Two-Port Networks Consider the noisy two-port network shown in Figure 2.13(a). Ys = Gs + j Bs is the source admittance connected at port 1 of the network. The noise it generates is represented by the current source is with a root-mean-square value of Is . This noisy two-port can be replaced by a noise-free network with a current source in and a voltage source vn connected at its input, as shown in Figure 2.13(b). In and Vn represent the corresponding root-mean-square current and voltage of the noise. It is assumed that the noise represented by is is uncorrelated with that represented by in and vn . However, a part of in , inc , is assumed to be correlated with vn via the correlation admittance Yc = Gc + j Xc while the remaining part inu is uncorrelated. Hence, Is2 = is2 = 4kTBG S Vn2

=

vn2

(2.5.12)

= 4kTBR n

(2.5.13)

and 2 2 = inu = 4kTBG nu Inu

(2.5.14)

I1

↑

Is

Ys

I2

Noisy two-port network

V1

V2

(a) I1

Is

↑

Ys

I2

Vn

V1

In

↑

Noise-free two-port network

(b)

Figure 2.13 Noisy two-port network (a) and its equivalent circuit (b).

V2

40

COMMUNICATION SYSTEMS

Now we find a Norton equivalent for the circuit that is connected at the input of the noise-free two-port network shown in Figure 2.13(b). Since these are random 2 variables, the mean-square noise current < ieq > is found as follows: 2 = is2 + |in + Ys vn |2 = is2 + |inu + inc + Ys vn |2 ieq

= is2 + |inu + (Yc + Ys )vn |2 or 2 2 ieq = is2 + inu + |Yc + Yn |2 vn2

(2.5.15)

Hence, the noise factor F is F =

2 ieq

is2

=1+

2 v 2 Rn inu Gnu + |Yc + Ys |2 + |Yc + Ys |2 2n = 1 + 2 is is Gs Gs

or F =1+

Gnu Rn + [(Gs + Gc )2 + (Xs + Xc )2 ] Gs Gs

(2.5.16)

For a minimum noise factor, Fmin , ∂F Gnu = 0 ⇒ G2s = G2c + = G2opt ∂Gs Rn

(2.5.17)

∂F = 0 ⇒ Xs = −Xc = Xopt ∂Xs

(2.5.18)

and

From (2.5.17), Gnu = Rn (G2opt − G2c )

(2.5.19)

Substituting (2.5.18) and (2.5.19) into (2.5.16), the minimum noise factor is found as Fmin = 1 + 2Rn (Gopt + Gc ) (2.5.20) Using (2.5.18) to (2.5.20), (2.5.16) can be expressed as F = Fmin +

Rn |Ys − Yopt |2 Gs

(2.5.21)

Noise Figure of a Cascaded System Consider a two-port network with gain G1 , noise factor F1 , and equivalent noise temperature Te1 . It is connected in cascade with another two-port network, as shown in Figure 2.14(a). The second two-port network has gain G2 , noise factor

41

NOISE AND DISTORTION

Ni

N1

G1, F1, Te1

Ti

G2, F2, Te2

No

(a) Ni

No

G, F, Te

Ti

(b)

Figure 2.14

Two networks connected in cascade (a) and its equivalent system (b).

F2 , and noise temperature Te2 . Our goal is to find the noise factor F and equivalent noise temperature Te of the overall system illustrated in Figure 2.14(b). Assume that the noise power input to the first two-port network is Ni . Its equivalent noise temperature is Ti . The output noise power of the first system is N1 , whereas it is No after the second system. Hence, N1 = G1 kTi B + G1 kTe1 B

(2.5.22)

and No = G2 N1 + G2 kTe2 B = G1 G2 kB(Ti + Te1 + Te2 /G1 ) = G2 [G1 kB(Ti + Te1 )] + G2 kTe2 B or No = G1 G2 kB(Te + Ti ) = GkB(Te + Ti )

(2.5.23)

Therefore, the noise temperature of a cascaded system is Te = Te1 +

Te2 G1

and from Te1 = (F1 − 1)Ti , Te2 = (F2 − 1)Ti , and Te = (F − 1)Ti , we get F = F1 +

F2 − 1 G1

The equations for Te and F above can be generalized as follows: Te2 Te3 Te4 + + + ··· G1 G1 G2 G1 G2 G3

(2.5.24)

F2 − 1 F3 − 1 F4 − 1 + + +··· G1 G1 G2 G1 G2 G3

(2.5.25)

Te = Te1 + and F = F1 +

42

COMMUNICATION SYSTEMS

Example 2.11 A receiving antenna is connected to an amplifier through a transmission line that has an attenuation of 2 dB. The gain of the amplifier is 15 dB, and its noise temperature is 150 K over a bandwidth of 100 MHz. All the components are at an ambient temperature of 300 K. (a) Find the noise figure of this cascaded system. (b) What would be the noise figure if the amplifier were placed before the transmission line? SOLUTION First we need to determine the noise factor of the transmission line alone. The formulas derived in the preceding section can then provide the noise figures desired for the two cases. Consider a transmission line that is matchterminated at both its ends by resistors R0 , as illustrated in Figure 2.15. Since the entire system is in thermal equilibrium at T (K), noise powers delivered to the transmission line and available at its output are kTB. Mathematically, Po = kTB = GkTB + GN added where Nadded is noise generated by the line as it appears at its input terminals. G is an output-to-input power ratio and B is the bandwidth of the transmission line. Note that input noise is attenuated in a lossy transmission line, but there is noise generated by it as well. Hence, 1 1 Nadded = (1 − G)kTB = − 1 kTB = kTe B G G An expression for the equivalent noise temperature Te of the transmission line is found from it as follows: 1 −1 T Te = G and F =1+

Te =1+ T0

1 T −1 G T0

The gain of the amplifier, Gamp = 15 dB = 101.5 = 31.6228. For the transmission line, 1/G = 100.2 = 1.5849. Hence, the noise factor of the line is 1 + (1.5849 − 1)300/290 = 1.6051. The corresponding noise figure is 2.05 dB. Similarly, the

kTB T R0

G, T, R0

Po = kTB

T R0

Figure 2.15 Lossy transmission line that is matched terminated at its ends.

43

NOISE AND DISTORTION

noise factor of the amplifier is found to be 1 + 150/290 = 1.5172. Its noise figure is 1.81 dB. (a) In this case the noise figure of the cascaded system, Fcascaded , is found as Famp − 1 1.5172 − 1 = 2.4248 = 3.8468 dB = 1.6051 + Gline 1/1.5849

Fcascaded = Fline +

(b) If the amplifier is connected before the line (i.e., the amplifier is placed right at the antenna), Fcascaded = Famp +

Fline − 1 1.6051 − 1 = 1.5363 = 1.8649 dB = 1.5172 + Gamp 31.6228

Note that the amplifier alone has a noise figure of 1.81 dB. Hence, the noisy transmission line connected after it does not alter the noise figure significantly. Example 2.12 Two amplifiers, each with a 20-dB gain, are connected in cascade as shown in Figure 2.16. The noise figure of amplifier A1 is 3 dB, while that of A2 is 5 dB. Calculate the overall gain and noise figure for this arrangement. If the order of two amplifiers is changed in the system, find its resulting noise figure. SOLUTION The noise factors and gains of two amplifiers are F1 = 3 dB = 100.3 = 2 F2 = 5 dB = 100.5 = 3.1623 G1 = G2 = 20 dB = 102 = 100 Therefore, the overall gain and noise figure of the cascaded system is found as follows: G=

P3 P3 P2 = × = 100 × 100 = 10, 000 = 40 dB P1 P2 P1

and F = F1 +

F2 − 1 3.1623 − 1 =2+ = 2.021623 = 3.057 dB G1 100

P1

P2 A1

Figure 2.16

P3 A2

Setup for Example 2.12.

44

COMMUNICATION SYSTEMS

If the order of amplifiers is changed, the overall gain will stay the same. However, the noise figure of the new arrangement will change as follows: F = F2 +

F1 − 1 2−1 = 3.1623 + = 3.1723 = 5.013743 dB G2 100

Minimum Detectable Signal Consider a receiver circuit with gain G over a bandwidth B. Assume that its noise factor is F . PI and Po represent the signal power at its input and output ports, respectively. NI is the input noise power and No is the total noise power at its output, as illustrated in Figure 2.17. Hence, No = kT0 F BG

(2.5.26)

This constitutes the noise floor of the receiver. A signal weaker than this will be lost in noise. No can be expressed in dBW as follows: No (dBW) = 10 log10 kTo + F (dB) + 10 log10 B + G(dB)

(2.5.27)

The minimum detectable signal must have power higher than this. Generally, it is taken as 3 dB above this noise floor. Further, 10 log10 kT0 = 10 log10 (1.38 × 10−23 × 290) ≈ −204 dBW/Hz Hence, the minimum detectable signal PoMDS at the output is PoMDS = −201 + F (dB) + 10 log10 BHz + G(dB)

(2.5.28)

The corresponding signal power PI MDS at its input is PI MDS (dBW) = −201 + F (dB) + 10 log10 BHz

(2.5.29)

Alternatively, PI MDS can be expressed in dBm as follows: PI MDS (dBm) = −111 + F (dB) + 10 log10 BMHz

NI, PI

Receiver

(2.5.30)

No , Po

Figure 2.17 Signals at the two ports of a receiver with a noise figure of F decibels.

45

NOISE AND DISTORTION

Example 2.13 The noise figure of a communication receiver is found as 10 dB at room temperature (290 K). Determine the minimum detectable signal power if (a) B = 1 MHz, (b) B = 1 GHz, (c) B = 10 GHz, and (d) B = 1 kHz. SOLUTION From (2.5.30), PI MDS = −111 + F (dB) + 10 log10 BMHz dBm Hence, (a) PI MDS = −111 + 10 + 10 log10 (1) = −101 dBm = 7.94 × 10−11 mW (b) PI MDS = −111 + 10 + 10 log10 (103 ) = −71 dBm = 7.94 × 10−8 mW (c) PI MDS = −111 + 10 + 10 log10 (104 ) = −61 dBm = 7.94 × 10−7 mW (d) PI MDS = −111 + 10 + 10 log10 (10−3 ) = −131 dBm = 7.94 × 10−14 mW These results show that the receiver can detect a relatively weak signal when its bandwidth is narrow. Intermodulation Distortion The electrical noise of a system determines the minimum signal level that it can detect. On the other hand, the signal will be distorted if its level is too high. This occurs because of the nonlinear characteristics of electrical devices, such as diodes, transistors, and so on. In this section we analyze the distortion characteristics and introduce the associated terminology. Consider the nonlinear system illustrated in Figure 2.18. Assume that its nonlinearity is frequency independent and can be represented by the following power series: (2.5.31) vo = k1 vi + k2 vi2 + k3 vi3 + · · · For simplicity, we assume that the ki are real and the first three terms of this series are sufficient to represent its output signal. Further, it is assumed that the input signal has two different frequency components that can be expressed as follows: (2.5.32) vi = a cos ω1 t + b cos ω2 t Therefore, the corresponding output signal can be written as vo ≈ k1 (a cos ω1 t + b cos ω2 t) + k2 (a cos ω1 t + b cos ω2 t)2 + k3 (a cos ω1 t + b cos ω2 t)3

ni

Figure 2.18

Nonlinear circuit

no

Nonlinear circuits with input signal vi that produces vo at its output.

46

COMMUNICATION SYSTEMS

After simplifying and rearranging it, we get

b2 a2 (1 + cos 2ω1 t) + (1 + cos 2ω2 t) 2 2 ab + {cos(ω1 + ω2 )t + cos(ω1 − ω2 t)} 2 3 3 3 2 a3 4 a cos ω1 t + 2 ab cos ω1 t + 4 cos 3ω1 t 3 2 3 2 + ab cos(ω1 − 2ω2 )t + a b cos(2ω1 − ω2 )t 4 4 (2.5.33) + k3 3 2 3 3 b3 + a b cos ω2 t + b cos ω2 t + cos 3ω2 t 2 4 4 3 2 3 2 + a b cos(2ω1 + ω2 )t + ab cos(ω1 + 2ω2 )t 4 4

vo = k1 (a cos ω1 t + b cos ω2 t) + k2

Therefore, the output signal has several frequency components in its spectrum. Amplitudes of various components are listed in Table 2.6. Figure 2.19 illustrates the input–output characteristic of an amplifier. If the input signal is too low, it may be submerged under the noise. The output power rises linearly above the noise as the input is increased. However, it deviates from the linear characteristic after a certain level of input power. In the linear region, output power can be expressed in dBm as follows: Pout (dBm) = Pin (dBm) + G(dB) The input power for which output deviates by 1 dB below its linear characteristic is known as 1-dB compression point. In this figure, it occurs at an input power

1 dB

Po1

Pout(dBm)

Noise level 1- dB compression point

Dynamic range PI MDS

Figure 2.19

Pin(dBm)

PD

Gain characteristics of an amplifier.

NOISE AND DISTORTION

47

TABLE 2.6 Amplitudes of Various Harmonics in the Output Harmonic Components ω1 ω2 ω1 − ω2 ω1 + ω2 2ω1 2ω2 3ω1 3ω2 2ω1 − ω2 ω1 − 2ω2 2ω1 + ω2 ω1 + 2ω2

Amplitude 3 3 3 2 a + ab k1 a + k3 4 2 3 3 3 2 b + a b k1 b + k3 4 2 ab k2 2 ab k2 2 a2 k2 2 b2 k2 2 a3 k3 4 b3 k3 4 3 k3 a 2 b 4 3 k3 ab2 4 3 k3 a 2 b 4 3 k3 ab2 4

of PD (dBm) that produces an output of Po1 (dBm). From the relation above, we find that Po1 (dBm) + 1 = PD (dBm) + G(dB) or PD (dBm) = Po1 (dBm) + 1 − G(dB)

(2.5.34)

The difference between the input power at 1-dB compression point and the minimum detectable signal defines the dynamic range (DR). Hence, DR = PD (dBm) − PI MDS From (2.5.30) and (2.5.34), we find that DR = Po1 (dBm) + 112 − G(dB) − F (dB) − 10 log10 BMHz

(2.5.35)

48

COMMUNICATION SYSTEMS

Gain Compression Nonlinear characteristics of the circuit (amplifier, mixer, etc.) compress its gain. If there is only one input signal [i.e., b is zero in (2.5.32)], the amplitude a1 of cos ω1 t in its output is found from Table 2.6 to be a1 = k1 a + 34 k3 a 3

(2.5.36)

The first term of a1 represents the linear (ideal) case, while its second term results from the nonlinearity. At the 1-dB compression point, k1 a + 34 k3 a 3 k1 a + 34 k3 a 3 dB = k1 a|dB − 1 → 20 log k1 a = −1 = −20 log (1.122) or

k1 + 34 k3 a 2 k1 + 34 k3 a 2 × 1.122 = 0 → × 1.122 = 1 20 log k1 k1

Therefore, k3 = −0.145

k1 a2

(2.5.37)

This indicates that k3 is a negative constant for a positive k1 , or vice versa. Therefore, it tends to reduce a1 , resulting in a lower gain. The single-tone gain compression factor may be defined as follows: A1C =

a1 3k3 2 a =1+ k1 a 4k1

(2.5.38)

Let us now consider the case when both of the input signals are present in (2.5.32). The amplitude of cos ω1 t in the output now becomes k1 a + k3 34 a 3 + 32 ab2 . If b is large compared with a, the term with k3 may dominate (undesired) over the first (desired) one. Since k3 and k1 have opposite signs, the desired output k1 a may be blocked completely for a certain value of b. Second Harmonic Distortion Second harmonic distortion occurs due to k2 . If b is zero, the amplitude of the a2 second harmonic will be k2 . Since power is proportional to the square of 2 the voltage, the desired term in the output can be expressed as a 2 = 20 log10 a + C1 P1 = 10 log10 ζk1 2

(2.5.39)

49

NOISE AND DISTORTION

where ζ is the proportionality constant and C1 = 20 log

ζk1 2

Similarly, power in the second harmonic component can be expressed as P2 = 10 log10

a2 ςk2 2

2 = 40 log10 a + C2

(2.5.40)

and the input power is a 2 = 20 log10 a + C3 Pin = 10 log10 ς 2

(2.5.41)

Proportionality constants ς and k2 are embedded in C2 and C3 . From (2.5.39)– (2.5.41) we find that P1 = Pin + D1

(2.5.42)

P2 = 2Pin + D2

(2.5.43)

and

where D1 and D2 replace C1 − C3 and C2 − 2C3 , respectively. Equations (2.5.42) and (2.5.43) indicate that both the fundamental and second harmonic signals in the output are linearly related with input power. However, the second harmonic power increases at twice the rate of the fundamental (the desired) component.

Intermodulation Distortion Ratio From Table 2.6 we find that the cubic term produces intermodulation frequencies 2ω1 ± ω2 and 2ω2 ± ω1 . If ω1 and ω2 are very close, 2ω1 + ω2 and 2ω2 + ω1 will be far away from the desired signals, and therefore, these can be filtered out easily. However, the other two terms, 2ω1 − ω2 and 2ω2 − ω1 , will be so close to ω1 and ω2 that these components may be within the passband of the system. It will distort the output. This characteristic of a nonlinear circuit is specified via the intermodulation distortion. It is obtained after dividing the amplitude of one of the intermodulation terms by the desired output signal. For an input signal with both ω1 and ω2 (i.e., a two-tone input), the intermodulation distortion ratio (IMR) may be found as IMR =

3 k a2b 4 3

k1 a

=

3k3 ab 4k1

(2.5.44)

50

COMMUNICATION SYSTEMS

Intercept Point Since power is proportional to the square of the voltage, the intermodulation distortion power may be defined as 3 PIMD = ς

ka 4 3

2

b

2 (2.5.45)

2

where ς is the proportionality constant. If the two input signals are equal in amplitude, a = b and the expression for intermodulation distortion power simplifies to 3 2 k a3 4 3 PIMD = ς (2.5.46) 2 Similarly, the power in one of the input signal components can be expressed as Pin = ς

(a)2 2

Therefore, the intermodulation distortion power can be expressed as PIMD = αPin3 where α is another constant. The ratio of intermodulation distortion power (PIMD ) to the desired output power Po (Po = k12 Pin ) for the case where two input signal amplitudes are the same is known as the intermodulation distortion ratio. It is found to be PIMR =

PIMD = α1 Pin2 Po

(2.5.47)

Note that PIMD increases as the cube of input power and the desired signal power Po is linearly related with Pin . Hence, PIMD increases three times as fast as Po on a log-log plot (or both of them are expressed in dBm before displaying on a linear graph). In other words, for a change of 1 dBm in Pin , Po changes by 1 dBm, whereas PIMD changes by 3 dBm. The value of the input power for which PIMD is equal to Po is referred to as the intercept point (IP). It is also referred to as IIP3 and the corresponding output power as OIP3 . Hence, PIMR is unity at the intercept point. If PIP is input power at IP, PIMR = 1 = α1 PIP2 ⇒ α1 =

1 PIP2

(2.5.48)

Therefore, the intermodulation distortion ratio PIMR is related to PIP as PIMR =

Pin PIP

2 (2.5.49)

51

NOISE AND DISTORTION

Example 2.14 The intercept point in the transfer characteristic of a nonlinear system is found to be at 25 dBm. If a −15-dBm signal is applied to this system, find the intermodulation ratio. SOLUTION PIMR =

Pin PIP

2 → PIMR (dB) = 2[Pin (dBm) − PIP (dBm)] = 2(−15 − 25) = −80 dB

Dynamic Range As mentioned earlier, noise at one end and distortion at the other limit the range of detectable signals of a system. The amount of distortion that can be tolerated depends somewhat on the type of application. If we set the upper limit that a system can detect as the signal level at which intermodulation distortion is equal to the minimum detectable signal, we can formulate an expression for its dynamic range. Thus, the ratio of the signal power that causes distortion power (in one frequency component) to be equal to the noise floor and the minimum detectable signal is the dynamic range (DR) of the system (amplifier, mixer, or receiver). It is also known as the spurious-free dynamic range (SFDR). Since the ideal power output Po is linearly related to input as Po = k12 Pin

(2.5.50)

the distortion power with reference to input can be expressed as Pdi =

PIMD k12

(2.5.51)

Therefore, PIMR =

k 2 Pdi PIMD Pdi = 12 = = Po Pin k1 Pin

Pin PIP

2 (2.5.52)

If Pdi = Nf (noise floor at the input), Nf = Pin

Pin PIP

2 ⇒ Pin3 = PIP2 Nf ⇒ Pin = (PIP2 Nf )1/3

(2.5.53)

and the dynamic range is (P 2 Nf )1/3 = DR = IP Nf

PIP Nf

2/3 (2.5.54)

or DR(dB) = 23 [PIP (dBm) − Nf (dBm)]

(2.5.55)

52

COMMUNICATION SYSTEMS

Example 2.15 A receiver is operating at 900 MHz with its bandwidth at 500 kHz and the noise figure at 8 dB. If its input impedance is 50 and IP is 10 dBm, find its dynamic range. SOLUTION Nf = kT0 BF ⇒ Nf (dBm) = 10 log10 (1.38 × 10−23 × 290 × 500 × 103 × 103 ) + 8 = −108.99 dBm Therefore, DR = 23 (10 + 108.99) = 79.32 dB

SUGGESTED READING Balanis, C. A. , Antenna Theory. New York: Wiley, 1997. Cheah, J. Y. C., Introduction to wireless communications applications and circuit design, in L. E. Larsen (ed.), RF and Microwave Circuit Design for Wireless Communications. Boston: Artech House, 1996. Razavi, B., RF Microelectronics. Upper Saddle River, NJ: Prentice Hall, 1998. Schiller, J. H., Mobile Communication. Reading, MA: Addison-Wesley, 2000. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1997. Stutzman, W. L., and G. A. Thiele, Antenna Theory and Design. New York: Wiley, 1998.

PROBLEMS 2.1. (a) What is the gain in decibels of an amplifier with a power gain of 4? (b) What is the power gain ratio of a 5-dB amplifier? (c) Express 2 kW of power in terms of dBm and dBw. 2.2. The maximum radiation intensity of a 90% efficient antenna is 200 mW per unit solid angle. Find the directivity and gain (dimensionless and in dB) when (a) the input power is 40π milliwatts and (b) the radiated power is 40π milliwatts. 2.3. The normalized far-zone field pattern of an antenna is given by sin θ cos2 φ E= 0

0 θ π and 0 φ π/2, 3π/2 φ 2π elsewhere

Find the directivity of this antenna.

PROBLEMS

53

2.4. The radiation intensity pattern of an antenna is given by U (θ, φ) =

2.5.

2.6.

2.7.

2.8.

2.9.

10 sin θ cos2 φ 0 θ π and 0 φ π/2, 3π/2 φ 2π 0 elsewhere

Find the directivity of this antenna. For an X-band (8.2 to 12.4 GHz) rectangular antenna with aperture dimensions of 5.5 and 7.4 cm, find its maximum effective aperture (in cm2 ) when its gain is (a) 14.8 dB at 8.2 GHz, (b) 16.5 dB at 10.3 GHz, and (c) 18.0 dB at 12.4 GHz. Transmitting and receiving antennas operating at 1 GHz with gains of 20 and 15 dB, respectively, are separated by a distance of 1 km. Find the maximum power delivered to the load when the input power is 150 W, assuming that the antennas are polarization matched. An antenna with a total loss resistance of 1 is connected to a generator whose internal impedance is 50 + j 25. Assuming that the peak voltage of the generator is 2 V and the impedance of antenna is 74 + j 42.5, find the power (a) supplied by the source (real power), (b) radiated by the antenna, and (c) dissipated by the antenna. The antenna connected to a radio receiver induces 9 µV of root-meansquare voltage into its input impedance, which is 50 . Calculate the input power in watts, dBm, and dBW. If the signal is amplified by 127 dB before it is fed to an 8- speaker, find the output power. The electric field radiated by a rectangular aperture, mounted on an infinite ground plane with z perpendicular to the aperture, is given by E = θˆ cos φ − φˆ sin φ cos θ f (r, θ, φ)

where f (r, θ, φ) is a scalar function that describes the field variation of the antenna. Assuming that the receiving antenna is linearly polarized along the x-axis, find the polarization loss factor. 2.10. A radar receiver has a sensitivity of 10−12 W. If the radar antenna’s effective aperture is 1 m2 and the wavelength is 10 cm, find the transmitter power required to detect an object at a distance of 3 km with a radar cross section of 5 m2 . 2.11. Two space vehicles are separated by 108 m. Each has an antenna with D = 1000 operating at 2.5 GHz. If vehicle A’s receiver requires 1 pW for a 20-dB signal-to-noise ratio, what transmitter power is required on vehicle B to achieve this signal-to-noise ratio? 2.12. (a) Design an Earth-based radar system that receives 10−15 W of peak echo power from Venus. It is to operate at 10 GHz with a single antenna to be used for both transmitting and receiving. Specify the effective aperture of the antenna and the peak transmitter power. Assume that the Earth to

54

COMMUNICATION SYSTEMS

Venus distance is 3 light-minutes; the diameter of Venus is 13 × 106 m; and the radar cross section of Venus is 15% of its physical cross section. (b) If the system of part (a) is used to observe the moon, determine the power received. Assume that the moon diameter is 3.5 × 106 m, its radar cross section is 15% of the physical cross section, and the Earthto-moon distance is 1.2 light-seconds. 2.13. Consider an imaging satellite that sends close-up pictures of the planet Neptune. It uses a 10-W transmitter operating at 18 GHz and a 2.5-mdiameter parabolic dish antenna. What Earth-station system temperature is required to provide a signal-to-noise ratio of 3 dB for reception of a picture with 3 × 106 pixels (picture elements) in 2 min if the Earth-station antenna diameter is 75 m? Assume aperture efficiencies of 70% and the Earth–Neptune distance as 4 light-hours. One pixel is equal to one bit, and two bits per second has a bandwidth of 1 Hz. 2.14. A Doppler radar operating at 10 GHz employs an antenna with its gain at 20 dB. If the signal reflected back from a target is a uniform plane wave with a power density of 10−8 W/m2 at the antenna, find the power delivered to a matched receiver. Also, if this radar is intended to detect target velocities ranging from 90 to 130 km/h, find the required passband of the Doppler filter. 2.15. A satellite transmitter is 36,500 km away from the receiver. It radiates a power of 10 W through an antenna of 19 dB gain toward the receiver. The satellite operates at a frequency of 11.5 GHz. If the receiving antenna has a gain of 60.5 dB, find the received power in dBm. 2.16. Consider the WLAN receiver front end shown in Figure P2.16, in which the bandpass filter has a bandwidth of 250 MHz centered at 5.4 GHz. If the system is operating at room temperature (290 K), calculate (a) the overall noise figure in decibels and (b) the minimum detectable signal in milliwatts. (c) Suggest if there is a better arrangement to increase the signal to noise ratio at the output.

Pin = −90 dBm

BPF

Pout

G = −1.5 dB

G = 10 dB F = 2 dB

G = 15 dB F = 2 dB

Figure P2.16

2.17. Various units of a 4-GHz receiver have the gains and noise temperatures indicated in Figure P2.17. (a) If the transmission line has a loss of 2 dB and the system is at an ambient temperature of 300 K, find the noise figure (in decibels) of this receiver.

55

PROBLEMS

RF amplifier

Mixer

IF amplifier

TRF = 50 K GRF = 50 dB

Tm = 500 K Gm = −10 dB

TIF = 1000 K GIF = 30 dB

Transmission line

Figure P2.17

(b) Suggest a better receiver architecture (if it exists) that improves the noise figure. Also, find the noise temperature of the new arrangement. 2.18. Two receivers for a design trade-off study are shown in Figure P2.18(a) and (b).

Mixer

IF amplifier

12−14 GHz

IF amplifier 2− 4 GHz

BPF

10 GHz

F = 6 dB G = −6 dB

F = 2 dB G = −2 dB

F = 5 dB G = 10 dB

F = 2 dB G = 15 dB

LO (a)

12−14 GHz

RF amplifier

F = 5 dB G = 10 dB

Mixer

10 GHz

IF amplifier F = 6 dB G = −6 dB

BPF F = 2 dB G = −2 dB

2 −4 GHz F = 4 dB G = 15 dB

LO (b)

Figure P2.18

(a) Calculate the noise figure for the two systems. (b) Calculate the overall gain for the two systems. (c) Calculate the minimum detectable signal at the input for the two systems. (d) If the output power at the 1-dB compression point is 10 mW for both systems, calculate the dynamic range for the two systems. 2.19. Calculate the input minimum detectable signal in milliwatts at room temperature for a receiver with (a) BW = 1 GHz, F = 5 dB and (b) BW = 100 MHz, F = 10 dB. 2.20. (a) What is the root-mean-square (rms) noise voltage produced in a 10-k resistance when its temperature is 45◦ C and the effective bandwidth is 100 MHz?

56

COMMUNICATION SYSTEMS

(b) A 50-k resistance is connected in parallel with the 10-k resistance of part (a). What is the resulting rms noise voltage? 2.21. The intercept point in the transfer characteristic of a nonlinear system is found to be 33 dBm. If a −18-dBm signal is applied to this system, find the intermodulation ratio. 2.22. A receiver is operating at 2455 MHz with its bandwidth at 500 kHz and the noise figure at 15 dB. If its input impedance is 50 and IP is 18 dBm, find its dynamic range. 2.23. An amplifier has a gain of 6 dB. Its intermodulation ratio is found to be −40 dB at an input power level of −4 dBm. Calculate the values of its input and output intercept points. What input signal level is needed if we want an intermodulation ratio of −50 dB?

3 TRANSMISSION LINES

Transmission lines are needed for connecting various circuit elements and systems. Open wire and coaxial lines are commonly used for circuits operating at low frequencies. On the other hand, coaxial line, strip line, microstrip line, and waveguides are employed at radio and microwave frequencies. Generally, lowfrequency signal characteristics are not affected as the signal propagates through the line. However, radio frequency and microwave signals are affected significantly because of the fact that the circuit size is comparable to the wavelength. A comprehensive understanding of signal propagation requires the analysis of electromagnetic fields in a given line. On the other hand, a generalized formulation can be obtained using circuit concepts on the basis of line parameters. This chapter begins with an introduction to line parameters and a distributed model of the transmission line. Solutions to the transmission line equation are then constructed to explain the behavior of the propagating signal. This is followed by discussions on sending-end impedance, reflection coefficient, return loss, and insertion loss. A quarter-wave impedance transformer is also presented, along with a few examples of matching resistive loads. Impedance measurement using the voltage standing wave ratio is then discussed. Finally, the Smith chart is introduced to facilitate graphical analysis and design of transmission line circuits. 3.1 DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES Any transmission line can be represented by a distributed electrical network (Figure 3.1) comprised of series inductors and resistors and shunt capacitors and Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

57

58

TRANSMISSION LINES R ∆z L ∆z

R ∆z

R ∆z

L ∆z C ∆z

G ∆z

L ∆z C ∆z

G ∆z

C ∆z

G ∆z

∆z

Figure 3.1 Distributed network model of a transmission line.

resistors. These distributed elements are defined as follows: L = inductance per unit length (H/m) R = resistance per unit length (/m) C = capacitance per unit length (F/m) G = conductance per unit length (S/m) L, R, C, and G, the line parameters, are determined theoretically by electromagnetic field analysis of a transmission line. These parameters are influenced by their cross-section geometry and the electrical characteristics of their constituents. For example, if a line is made up of an ideal dielectric and a perfect conductor, its R and G values will be zero. If it is a coaxial cable with inner and outer radii a and b, respectively, as shown in Figure 3.2, then C=

55.63εr ln(b/a)

pF/m

(3.1.1)

and L = 200 ln(b/a)

nH/m

a

εr

b

Figure 3.2 Coaxial line geometry.

(3.1.2)

59

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

where εr is the dielectric constant of the material between two coaxial conductors of the line. If the coaxial line has small losses due to an imperfect conductor and insulator, its resistance and conductance parameters can be calculated as follows: 1 1 fGHz R ≈ 10 + /m (3.1.3) a b σ and G=

0.3495εr fGHz tan δ ln(b/a)

S/m

(3.1.4)

where tanδ is the loss tangent of the dielectric material, σ the conductivity (in S/m) of the conductors, and fGHz is the signal frequency in gigahertz. Characteristic Impedance of a Transmission Line Consider a transmission line that extends to infinity, as shown in Figure 3.3. The voltages and currents at several points on the line are as indicated. When a voltage is divided by the current through that point, the ratio is found to remain constant. This ratio is called the characteristic impedance of the transmission line. Mathematically, characteristic impedance = Z0 = V1 /I1 = V2 /I2 = V3 /I3 = · · · = Vn /In In actual electrical circuits, the length of the transmission lines is always finite. Hence, it seems that the characteristic impedance has no significance in the real world. However, that is not the case. When the line extends to infinity, an electrical signal continues propagating in the forward direction without reflection. On the other hand, it may be reflected back by the load that terminates a transmission line of finite length. If one varies this termination, the strength of the reflected signal changes. When the transmission line is terminated by a load impedance that absorbs all the incident signal, the voltage source sees an infinite electrical length. The voltage-to-current ratio at any point on this line is a constant equal

I1

I2

V1

V2

I3

In

V3

Vn

Figure 3.3 Infinitely long transmission line and voltage source.

∞

60

TRANSMISSION LINES

to the terminating impedance. In other words, there is a unique impedance for every transmission line that does not produce an echo signal when the line is terminated by it. The terminating impedance that does not produce an echo on the line is equal to its characteristic impedance. If Z = R + j ωL = impedance per unit length Y = G + j ωC = admittance per unit length then using the definition of characteristic impedance and the distributed model shown in Figure 3.1, we can write Z0 = =

(Z0 + Z z)(1/Y z) Z0 + Z z + 1/Y z Z0 + Z z ⇒ Z0 Y (Z0 + Z z) = Z 1 + Y z(Z0 + Z z)

For z → 0,

Z0 =

Z = Y

R + j ωL G + j ωC

(3.1.5)

Some special cases are as follows: √ 1. For a dc signal, Z0 dc = R/G. √ 2. For ω → ∞, ωL R and ωC G; therefore, Z0(ω→large) = L/C. √ 3. For a lossless line, R → 0 and G → 0; therefore, Z0 = L/C. Thus, a lossless semirigid coaxial line with 2a = 0.036 in., 2b = 0.119 in., and an εr value of 2.1(Teflon-filled) will have C = 97.71 pF/m and L = 239.12 nH/m. Its characteristic impedance will be 49.5 . Since the conductivity of copper is 5.8 × 107 S/m and the loss tangent of Teflon is 0.00015, Z = 3.74 + j 1.5 × 103 /m and Y = 0.092 + j 613.92 mS/m at 1 GHz. The corresponding characteristic impedance is 49.5 − j 0.058 , which is very close to the approximate value of 49.5 . Example 3.1 Calculate the equivalent impedance and admittance of a 1-mlong line that is operating at 1.6 GHz. The line parameters are L = 0.002 µ H/m, C = 0.012 pF/m, R = 0.015 /m, and G = 0.1 mS/m. What is the characteristic impedance of this line? SOLUTION Z = R + j ωL = 0.015 + j 2π × 1.6 × 109 × 0.002 × 10−6 /m = 0.015 + j 20.11 /m

61

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

Y = G + j ωC = 0.0001 + j 2π × 1.6 × 109 × 0.012 × 10−12 S/m Z0 =

= 0.1 + j 0.1206 mS/m Z = 337.02 + j 121.38 Y

Transmission Line Equations Consider the equivalent distributed circuit of a transmission line that is terminated by a load impedance ZL , as shown in Figure 3.4. The line is excited by a voltage source v(t) with its internal impedance ZS . We apply Kirchhoff’s voltage and current laws over a small length, z, of this line as follows: For the loop, v(z, t) = L z or

∂i(z, t) + R z i(z, t) + v(z + z, t) ∂t

∂i(z, t) v(z + z, t) − v(z, t) = −Ri(z, t) − L z ∂t

Under the limit z → 0, the equation above reduces to ∂i(z, t) ∂v(z, t) = − R × i(z, t) + L ∂z ∂t

(3.1.6)

Similarly, at node A, i(z, t) = i(z + z, t) + G z v(z + z, t) + C z

∂v(z + z, t) ∂t

or ∂v(z + z, t) i(z + z, t) − i(z, t) = − G × v(z + z, t) + C z ∂t

ZS

v (z + ∆z, t) i(z + ∆z, t) A

v(z,t) L ∆z

v(t)

R ∆z C ∆z

z

Figure 3.4

i(z,t) L ∆z R ∆z G ∆z

C ∆z

L ∆z G ∆z

R ∆z C ∆z

∆z

Distributed circuit model of a transmission line.

G ∆z

ZL

62

TRANSMISSION LINES

Again, under the limit z → 0, it reduces to ∂v(z, t) ∂i(z, t) = −Gv(z, t) − C ∂z ∂t

(3.1.7)

Now, from equations (3.1.6) and (3.1.7), v(z, t) or i(z, t) can be eliminated to formulate the following: ∂ 2 v(z, t) ∂v(z, t) ∂ 2 v(z, t) = RGv(z, t) + (RC + LG) + LC ∂z2 ∂t ∂t 2

(3.1.8)

∂i(z, t) ∂ 2 i(z, t) ∂ 2 i(z, t) = RGi(z, t) + (RC + LG) + LC ∂z2 ∂t ∂t 2

(3.1.9)

and

Some special cases are as follows: 1. For a lossless line, R and G will be zero, and these equations reduce to well-known homogeneous scalar wave equations, ∂ 2 v(z, t) ∂ 2 v(z, t) = LC ∂z2 ∂t 2

(3.1.10)

∂ 2 i(z, t) ∂ 2 i(z, t) = LC ∂z2 ∂t 2

(3.1.11)

and

√ It is to be noted that the velocity of these waves is 1/ LC. 2. If the source is sinusoidal with time (i.e., time-harmonic), we can switch to phasor voltages and currents. In that case, equations (3.1.8) and (3.1.9) can be simplified as follows: d 2 V (z) = ZY V (z) = γ2 V (z) dz2

(3.1.12)

d 2 I (z) = ZY I (z) = γ2 I (z) dz2

(3.1.13)

and

where V (z) and I (z) are phasor quantities; Z and Y are impedance per unit length √ and admittance per unit length, respectively, as defined earlier. γ = ZY = α + j β is known as the propagation constant of the line. α and β are called the attenuation constant and the phase constant, respectively. Equations (3.1.12) and (3.1.13) are referred to as homogeneous Helmholtz equations.

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

63

Solution of Helmholtz Equations Note that both differential equations have the same general format. Therefore, we consider the solution to the following generic equation here. Expressions for voltage and current on the line can be constructed on the basis of that. d 2 f (z) − γ2 f (z) = 0 dz2

(3.1.14)

Assume that f (z) = Ceκz , where C and κ are arbitrary constants. Substituting it into (3.1.14), we find that κ = ±γ. Therefore, a complete solution to this equation may be written as follows: f (z) = C1 e−γz + C2 eγz

(3.1.15)

where C1 and C2 are integration constants that are evaluated through the boundary conditions. Hence, complete solutions to equations (3.1.12) and (3.1.13) can be written as follows: V (z) = Vin e−γz + Vref eγz

(3.1.16)

I (z) = Iin e−γz + Iref eγz

(3.1.17)

and

where Vin , Vref , Iin , and Iref are integration constants that may be complex, in general. These constants can be evaluated from the known values of voltages and currents at two different locations on the transmission line. If we express the first two of these constants in polar form as Vin = vin ej φ

and Vref = vref ej ϕ

the line voltage, in time domain, can be evaluated as follows: v(z, t) = Re[V (z)ej ωt ] = Re[Vin e−(α+j β)z ej ωt + Vref e+(α+j β)z ej ωt ] or v(z, t) = vin e−αz cos(ωt − βz + φ) + vref e+αz cos(ωt + βz + ϕ) (3.1.18) At this point, it is important to analyze and understand the behavior of each term on the right-hand side of this equation. At a given time, the first term changes sinusoidally with distance, z, while its amplitude decreases exponentially. It is illustrated in Figure 3.5(a). On the other hand, the amplitude of the second sinusoidal term increases exponentially. It is shown in Figure 3.5(b). Further, the argument of the cosine function decreases with distance in the former case and

TRANSMISSION LINES 1.2

150.0

0.8

100.0

0.4

50.0

f2 (Z)

f1 (Z)

64

0.0 −0.4 −0.8 0.0

−50.0

λ 1.0

0.0

2.0

3.0

4.0

5.0

−100.0 0.0

λ 1.0

Z (a)

Figure 3.5

2.0

3.0

4.0

5.0

Z (b)

Behavior of two solutions to the Helmholtz equation with distance.

increases in the latter case. When a signal is propagating away from the source along +z-axis, its phase should be delayed. Further, if it is propagating in a lossy medium, its amplitude should decrease with distance z. Thus, the first term on the right-hand side of equation (3.1.16) represents a wave traveling along the +z-axis (an incident or outgoing wave). Similarly, the second term represents a wave traveling in the opposite direction (a reflected or incoming wave). This analysis is also applied to equation (3.1.17). Note that Iref is reflected current that will be 180◦ out of phase with incident current Iin . Hence, Vref Vin =− = Z0 Iin Iref and therefore equation (3.1.17) may be written as follows: I (z) =

Vin −γz Vref γz e − e Z0 Z0

(3.1.19)

Incident and reflected waves change sinusoidally with both space and time. Time duration over which the phase angle of a wave goes through a change of 360◦ (2π radians) is known as its time period. The inverse of the time period in seconds is the signal frequency in hertz. Similarly, the distance over which the phase angle of the wave changes by 360◦ (2π radians) is known as its wavelength (λ). Therefore, the phase constant β is equal to 2π divided by the wavelength in meters. Phase and Group Velocities The velocity with which the phase of a time-harmonic signal moves is known as its phase velocity. In other words, if we tag a phase point of the sinusoidal

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

65

wave and monitor its velocity, we obtain the phase velocity, vp , of this wave. Mathematically, ω vp = β A transmission line has no dispersion if the phase velocity of a propagating signal is independent of frequency. Hence, a graphical plot of ω versus β will be a straight line passing through the origin. This type of plot is called the dispersion diagram of a transmission line. An information-carrying signal is composed of many sinusoidal waves. If the line is dispersive, each of these harmonics will travel at a different velocity. Therefore, the information will be distorted at the receiving end. The velocity with which a group of waves travels is called the group velocity, vg . It is equal to the slope of the dispersion curve of the transmission line. Consider two sinusoidal signals with angular frequencies ω + δω and ω − δω, respectively. Assume that these waves of equal amplitudes are propagating in the z-direction with corresponding phase constants β + δβ and β − δβ. The resulting wave can be found as follows: f (z, t) = Re[Aej [(ω+δω)t−(β+δβ)z] + Aej [(ω−δω)t−(β−δβ)z] ] = 2A cos(δωt − δβz) cos(ωt − βz) Hence, the resulting wave, f (z, t), is amplitude modulated. The envelope of this signal moves with the group velocity, vg =

δω δβ

Example 3.2 A signal generator has an internal resistance of 50 and an opencircuit voltage v(t) = 3 cos(2π × 108 t) V. It is connected to a 75- lossless transmission line that is 4 m long and terminated by a matched load at the other end. If the signal propagation velocity on this line is 2.5 × 108 m/s, find the instantaneous voltage and current at an arbitrary location on the line. SOLUTION Since the transmission line is terminated by a load that is equal to its characteristic impedance, there will be no echo signal. Further, an equivalent circuit at its input end may be drawn as shown in Figure 3.6. Using the voltagedivision rule and Ohm’s law, incident voltage and current can be determined as follows: 75 ◦ ◦ 3 0 = 1.8 0 V 50 + 75 3 0◦ ◦ incident current at the input end, Iin (z = 0) = = 0.024 0 A 50 + 75

incident voltage at the input end, Vin (z = 0) =

66

TRANSMISSION LINES 50 Ω

50 Ω

4m

75 Ω

3V∠0°

75 Ω

75 Ω 3V∠0°

z

Figure 3.6

and β=

Circuit arrangement for Example 3.2.

2π × 108 ω = = 0.8π rad/m vp 2.5 × 108

Therefore, V (z) = 1.8e−j 0.8πz V and I (z) = 0.024e−j 0.8πz A. Hence, v(z, t) = 1.8 cos(2π × 108 t − 0.8πz) V and i(z, t) = 0.024 cos(2π × 108 t − 0.8πz) A Example 3.3 The parameters of a transmission line are R = 2 /m, G = 0.5 mS/m, L = 8 nH/m, and C = 0.23 pF/m. If the signal frequency is 1 GHz, calculate its characteristic impedance (Z0 ) and the propagation constant (γ). SOLUTION R + j ωL 2 + j 2π × 109 × 8 × 10−9 = Z0 = G + j ωC 0.5 × 10−3 + j 2π × 109 × 0.23 × 10−12 2 + j50.2655 50.31 1.531 rad = = −3 −3 0.5 × 10 + j 1.4451 × 10 15.29 × 10−4 1.2377 rad ◦

= 181.39 8.4 = 179.44 + j 26.51 and γ=

√ ZY = (50.31 1.531 rad) × (15.29 × 10−4 1.2377 rad) ◦

= 0.2774 79.31 m−1 = 0.0514 + j 0.2726 m−1 = α + j β Therefore, α = 0.0514 Np/m and β = 0.2726 rad/m. Example 3.4 Two antennas are connected through a quarter-wavelength-long lossless transmission line, as shown in Figure 3.7. However, the characteristic

67

SENDING-END IMPEDANCE

jX Z0 = ?

B

80 + j35 Ω

A

50 Ω 56 + j28 Ω

(λ/4)

z

Figure 3.7

Circuit arrangement for Example 3.4.

impedance of this line is unknown. The array is excited through a 50- line. Antenna A has an impedance of 80 + j 35 and antenna B has 56 + j 28 . Currents (peak values) through these antennas are found to be 1.5 0◦ and 1.5 90◦ A, respectively. Determine the characteristic impedance of the line connecting these two antennas and the value of a reactance connected in series with antenna B. SOLUTION Assume that Vin and Vref are the incident and reflected phasor voltages, respectively, at antenna A. Therefore, the current, IA , through this antenna is IA =

Vin − Vref ◦ ◦ = 1.5 0 A ⇒ Vin − Vref = Z0 IA = Z0 1.5 0 V Z0

Since the connecting transmission line is a quarter-wavelength long, the incident and reflected voltages across the transmission line at the location of B will be j Vin and −j Vref , respectively. Therefore, the total voltage, VTBX , appearing across the antenna B and the reactance jX combined will be equal to j (Vin − Vref ). ◦

VTBX = j (Vin − Vref ) = jZ0 IA = j 1.5Z0 = 1.5Z0 90 Also, Ohm’s law can be used to find this voltage as follows: ◦

VTBX = (56 + j 28 + j X)1.5 90

Therefore, X = −28 and Z0 = 56 . Note that the unknown characteristic impedance is a real quantity because the transmission line is lossless. 3.2 SENDING-END IMPEDANCE Consider a transmission line of length l and characteristic impedance Z0 . It is terminated by a load impedance ZL , as shown in Figure 3.8. Assume that the

68

TRANSMISSION LINES Iin Iref

Zin

z=0

Figure 3.8

Load ZL

Z0

V in + Vref

z=l

Transmission line terminated by a load impedance.

incident and reflected voltages at its input (z = 0) are Vin and Vref , respectively. The corresponding currents are represented by Iin and Iref . If V (z) represents the total phasor voltage at point z on the line and I (z) is the total current at that point, V (z) = Vin e−γz + Vref eγz

(3.2.1)

I (z) = Iin e−γz + Iref eγz

(3.2.2)

and

where Vin , Vref , Iin , and Iref are incident voltage, reflected voltage, incident current, and reflected current at z = 0, respectively. Impedance at the input of this transmission line, Zin , can be found after dividing total voltage by the total current at z = 0. Thus, Zin =

V (z = 0) Vin + Vref Vin + Vref Vin + Vref = = Z0 = I (z = 0) Iin + Iref Vin /Z0 − Vref /Z0 Vin − Vref

or Zin = Z0

1 + Vref /Vin 1 + 0 = Z0 1 − Vref /Vin 1 − 0

(3.2.3)

where 0 = ρej φ is known as the input reflection coefficient. Further, Zin = Z0

Zin 1 + 0 1 + 0 ⇒ = Z in = 1 − 0 Z0 1 − 0

where Z in is called the normalized input impedance. Similarly, voltage and current at z = l are related through load impedance as follows: ZL =

V (z = l) Vin e−γl + Vref e+γl = I (z = l) Iin e−γl + Iref e+γl

= Z0

Vin e−γl + Vref e+γl e−γl + 0 e+γl = Z 0 −γl Vin e−γl − Vref e+γl e − 0 e+γl

69

SENDING-END IMPEDANCE

Therefore, ZL =

e−γl + 0 eγl ZL − 1 −2γl ⇒ 0 = e e−γl − 0 e+γl ZL + 1

(3.2.4)

and equation (3.2.3) can be written as follows: Zin =

1 + [(ZL − 1)/(ZL + 1)]e−2γl 1 − [(ZL − 1)/(ZL + 1)]e−2γl

=

ZL (1 + e−2γl ) + (1 − e−2γl ) ZL (1 − e−2γl ) + (1 + e−2γl )

since 1 − e−2γl e+γl − e−γl sinh γl = tanh γl = +γl = −2γl 1+e e + e−γl cosh γl Z in =

Z L + tanh γl 1 + Z L tanh γl

or Zin = Z0

ZL + Z0 tanh γl Z0 + ZL tanh γl

(3.2.5)

For a lossless line, γ = α + j β = j β, and therefore tanh γl = tanh j βl = j tan βl. Hence, equation (3.2.5) simplifies as follows: Zin = Z0

ZL + j Z0 tan βl Z0 + j ZL tan βl

(3.2.6)

It is to be noted from this equation that Zin repeats periodically every one-half wavelength on the transmission line. In other words, the input impedance on a lossless transmission line will be the same at points d ± nλ/2, where n is an integer, due to the fact that βl =

2π nλ 2πd ± nπ d± = λ 2 λ

and tan βl = tan

2πd 2πd ± nπ = tan λ λ

Some special cases are as follows: 1. ZL = 0 (i.e., a lossless line is short circuited) ⇒ Zin = j Z0 tan βl. 2. ZL = ∞ (i.e., a lossless line has an open circuit at the load) ⇒ Zin = −j Z0 cot βl. 3. l = λ/4, and therefore βl = π/2 ⇒ Zin = Z02 /ZL .

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TRANSMISSION LINES

According to the first two cases, a lossless line can be used to synthesize an arbitrary reactance. The third case indicates that a quarter-wavelength-long line of suitable characteristic impedance can be used to transform a load impedance ZL to a new value of Zin . This type of transmission line, called an impedance transformer, is useful in impedance-matching applications. Further, this equation can be rearranged as follows: Z in =

1 ZL

= YL

Hence, normalized impedance at a point a quarter-wavelength away from the load is equal to the normalized load admittance. Example 3.5 A transmission line of length d and characteristic impedance Z0 acts as an impedance transformer to match a 150- load to a 300- line (see Figure 3.9). If the signal wavelength is 1 m, find (a) d, (b) Z0 , and (c) the reflection coefficient at the load. SOLUTION (a) d = λ/4 → d = 0.25 m √ (b) Z0 = Zin ZL → Z0 = (150 × 300)1/2 = 212.132 150 − 212.132 ZL − 1 ZL − Z0 = (c) L = = = 0.1716 ZL + Z0 150 + 212.132 ZL + 1 Example 3.6 Design a quarter-wavelength transformer to match a 20- load to the 45- line at 3 GHz. If this transformer is made from a Teflon-filled (εr = 2.1) coaxial line, calculate its length (in centimeters). Also, determine the diameter of its inner conductor if the inner diameter of the outer conductor is 0.5 cm. Assume that the impedance transformer is lossless. SOLUTION Z0 = and

√ Zin ZL = 45 × 20 = 30

Z0 =

L = C

b 200 × 10−9 ln = 30 −12 55.63 × εr × 10 a d

300 Ω Z0

Figure 3.9

Circuit arrangement for Example 3.5.

150 Ω

71

SENDING-END IMPEDANCE

Therefore, ln

b 30 = = 0.7251 a 41.3762

and 2b 0.5 2b b = = e0.7251 = 2.0649 ⇒ 2a = = = 0.2421 cm a 2a 2.0649 2.0649 √ Phase constant β = ω LC: 2π 1 ω√ LC = f × 200 × 55.63 × εr × 10−21 ⇒ = λ λ 2π √ 9 = 3 × 10 × 10−10 × 200 × 55.63 × 2.1 × 0.1

β=

= 14.5011 m−1 Therefore, λ = 0.06896 m and d = λ/4 = 0.01724 m = 1.724 cm. Example 3.7 Design a quarter-wavelength microstrip impedance transformer to match a patch antenna of 80 with a 50- line. The system is to be fabricated on a 1.6-mm-thick substrate (εr = 2.3) that operates at 2 GHz (see Figure 3.10). SOLUTION The characteristic impedance of the microstrip line impedance transformer must be Z0 = Z01 ZL = 63.2456 Design formulas for microstrip line are given in Appendix 2. Assume that the strip thickness t is less than 0.096 mm and that the dispersion is negligible for the time being at the operating frequency.

2.3 + 1 2

1/2

A=

63.2456 60

B=

60π2 = 6.1739 √ 63.2456 2.3

Z01

+

2.3 − 1 0.11 × 0.23 + = 1.4635 2.3 + 1 2.3

Z0

ZL

d

Figure 3.10 Circuit arrangement for Example 3.7.

72

TRANSMISSION LINES

Since A ≤ 1.52, 2 w = 6.1739 − 1 − ln(2 × 6.1739 − 1) h π 2.3 − 1 0.61 + ln(6.1739 − 1) + 0.39 − 2 × 2.3 2.3 =

2 × 3.2446 = 2.0656 ⇒ w = 3.3 mm π

At this point, we can check if the dispersion in the line is really negligible. For that, we determine the effective dielectric constant as follows: F

w h

−1/2 h −1/2 12 = 1 + 12 = 1+ = 0.383216 w 2.0656

Assuming that t/ h = 0.005, εe =

2.3 + 1 2.3 − 1 2.3 − 1 0.005 = 1.8981 ≈ 1.9 + × 0.383216 − ×√ 2 2 4.6 2.0656

and √

4h εr − 1 w 2 F = 0.5 + 1 + 2 × log 1 + = 0.213712 λ0 h √ 2 √ 2.3 − 1.9 √ + 1.9 = 1.909192 εe (f ) = 1 + 4 × F −1.5 Since εe (f ) is very close to εe , dispersion in the line can be neglected. Therefore, λ=

3 × 108 10.8821 = 2.72 cm √ m = 10.8821 cm ⇒ length of line = 9 4 2 × 10 × 1.9

Reflection Coefficient, Return Loss, and Insertion Loss The voltage reflection coefficient is defined as the ratio of reflected to incident phasor voltages at a location in the circuit. In the case of a transmission line terminated by load ZL , the voltage reflection coefficient is given by equation (3.2.4). Hence, =

Vref ZL − Z0 −2γl = e = ρL ej θ e−2(α+j β)l = ρL e−2αl e−j (2βl−θ) Vin ZL + Z0

(3.2.7)

73

SENDING-END IMPEDANCE 1.0

1.0

0.5 Imag. part

Imag. part

0.5

0.0

−0.5

−0.5 −1.0 −1.0

0.0

−0.5

0.0 Real part (a)

0.5

1.0

−1.0 −1.0

−0.5

0.0 Real part (b)

0.5

1.0

Figure 3.11 Reflection coefficient on (a) a lossy and (b) a lossless transmission line.

where ρL ej θ = (ZL − Z0 )/(ZL + Z0 ) is called the load reflection coefficient. Equation (3.2.7) indicates that the magnitude of reflection coefficient decreases by a factor of e−2αl as the observation point moves away from the load. Further, its phase angle changes by −2βl. A polar (magnitude and phase) plot of it will look like a spiral, as shown in Figure 3.11(a). However, the magnitude of reflection coefficient will not change if the line is lossless. Therefore, the reflection coefficient point will be moving clockwise on a circle of radius equal to its magnitude as the line length is increased. As illustrated in Figure 3.11 (b), it makes one complete revolution for each half-wavelength distance away from the load (because −2βλ/2 = −2π). Similarly, the current reflection coefficient, c , is defined as a ratio of reflected to incident signal-current phasors. It is related to the voltage reflection coefficient as follows: Iref −Vref /Z0 = = − c = Iin Vin /Z0 The return loss of a device is defined as the ratio of reflected power to incident power at its input. Since the power is proportional to the square of the voltage at that point, it may be found as return loss =

reflected power = ρ2 incident power

Generally, it is expressed in decibels as follows: return loss = 20 log10 ρ

dB

(3.2.8)

The insertion loss of a device is defined as the ratio of power transmitted (power available at the output port) to that of the power incident at its input. Since the power transmitted is equal to the difference of the incident and reflected

74

TRANSMISSION LINES

powers for a lossless device, the insertion loss can be expressed as follows: insertion lossof a lossless device = 10 log10 (1 − ρ2 )

dB

(3.2.9)

Low-Loss Transmission Lines Most practical transmission lines possess a very small loss of propagating signal. Therefore, expressions for the propagation constant and the characteristic impedance can be approximated for such lines as follows: γ=

√

ZY = (R + j ωL)(G + j ωC) =

−ω2 LC

R G 1+ 1+ j ωL j ωC

For R ωL and G ωC, a first-order approximation is √ γ = α + j β ≈ j ω LC 1 +

R G 1+ j 2ωL j 2ωC √ R G ≈ j ω LC 1 + + j 2ωL j 2ωC

Therefore, 1 C L R α≈ +G 2 L C and

√ β ≈ ω LC Z0 =

R + j ωL = G + j ωC

Np/m

rad/m

(3.2.10)

(3.2.11)

L G −1/2 R 1/2 1+ 1+ C j ωL j ωC

Hence,

L L R G R G Z0 ≈ 1+ 1− ≈ 1+ − C j 2ωL j 2ωC C j 2ωL j 2ωC L 1 R G 1+ − (3.2.12) = C j 2ω L C Thus, the attenuation constant of a low-loss line is independent of frequency, while its phase constant is linear, as in the case of a lossless line. However, the frequency dependency of its characteristic impedance is of concern to communication

75

SENDING-END IMPEDANCE

engineers, because it will distort the signal. If RC = GL, the frequency-dependent term will go to zero. This type of low-loss line is called a distortionless line. Hence, √ γ = ZY = (R + j ωL)(G + j ωC) RC C = α + jβ = (R + j ωL) ∵G = (3.2.13) L L Example 3.8 A signal propagating through a 50- distortionless transmission line attenuates at the rate of 0.01 dB/m. If this line has a capacitance of 100 pF/m, find (a) R, (b) L, (c) G, and (d) vp . SOLUTION Since the line is distortionless, L Z0 = = 50 C and

α = 0.01 dB/m ≈ 0.01/8.69 Np/m = 1.15 × 10−3 Np/m

Hence, (a) (b) (c) (d)

√ R = α L/C = 1.15 × 10−3 × 50 = 0.057 / m L = CZ02 = 10−10 × 502 H/m = 0.25 µH/m G = RC/L = R/Z02 = 0.057/502 S/m = 22.8µS/ m √ vp = 1/ LC = 2 × 108 m/s

Experimental Determination of Characteristic Impedance and Propagation Constant of a Transmission Line A transmission line of length d is kept open at one end, and the impedance at its other end is measured using an impedance bridge. Assume that it is Zoc . The process is repeated after placing a short circuit at its open end, and this impedance is recorded as Zsc . Using equation (3.2.5), one can write Zoc = Z0 coth γd

(3.2.14)

Zsc = Z0 tanh γd

(3.2.15)

and

Therefore, Z0 =

Zoc Zsc

and tanh γd =

(3.2.16)

1 Zsc ⇒ γ = tanh−1 Zoc d

Zsc Zoc

(3.2.17)

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TRANSMISSION LINES

These two equations can be solved to determine Z0 and γ. The following identity can be used to facilitate the evaluation of the propagation constant: tanh−1 Z =

1 1+Z ln 2 1−Z

The following examples illustrate this procedure. Example 3.9 Impedance at one end of the transmission line is measured to be Zin = 30 + j 60 using an impedance bridge, while its other end is terminated by a load ZL . The experiment is repeated twice, with the load replaced first by a short circuit and then by an open circuit. These data are recorded as j 53.1 and −j 48.3 , respectively. Find the characteristic resistance of this line and the load impedance. SOLUTION Z0 =

Zsc × Zoc = j 53.1 × (−j 48.3) = 50.6432

Zin = Z0 =

ZL + j Z0 tan βl ZL + Zsc = Z0 Z0 + j ZL tan βl Z0 + j ZL tan βl

Z0 ZL + Zsc j tan βl ZL + Z0 /(j tan βl)

Therefore, ZL + Zsc Zsc − Zin ⇒ ZL = Zoc ZL + Zoc Zin − Zoc j 53.1 − (30 + j 60) ZL = −j 48.3 × 30 + j 60 − (−j 48.3)

Zin = Zoc

Therefore, ZL = 11.6343 + j 6.3 Example 3.10 Measurements are made on a 1.5-m-long transmission line using an impedance bridge. After short-circuiting at one of its ends, impedance at the other end is found to be j 103 . Repeating the experiment with the short circuit now replaced by an open circuit gives −j 54.6 . Determine the propagation constant and the characteristic impedance of this line. SOLUTION Zoc Zsc = −j 54.6 × j 103 = 74.99 ≈ 75 j 103 tanh 1.5γ = = j 1.8969 ⇒ 1.5γ = tanh−1 j 1.8969 −j 54.3 Z0 =

77

SENDING-END IMPEDANCE

= 1.5γ =

1 1 + j 1.8969 ln 2 1 − j 1.8969 1 1 + j 1.8969 1 1 ln = ln(1 2.1713 rad) = ln(ej 2.1713 ) = j 1.08565 2 1 − j 1.8969 2 2

and γ = j 0.7238 m−1 . Example 3.11 A 10-m-long 50- lossless transmission line is terminated by a load, ZL = 100 + j 50 . It is driven by a signal generator that has an opencircuit voltage Vs at 100 0◦ V and source impedance Zs at 50 . The propagating signal has a phase velocity of 200 m/µS at 26 MHz. Determine the impedance at its input end and the phasor voltages at both its ends (see Figure 3.12). SOLUTION Zin = Z0 = 50

ZL + j Z0 tan βl Z0 + j ZL tan βl

and β =

ω 2π × 26 × 106 = = 0.8168 rad/m vp 200 × 106

100 + j 50 + j 50 tan(8.168) = 19.5278 0.181 rad 50 + j (100 + j 50) tan(8.168)

= 19.21 + j 3.52 The equivalent circuits at its input and at the load can be drawn as in Figure 3.13. Voltage at the input end, VA =

Vs 100 × 19.5278 0.181 rad Zin = Zs + Zin 50 + 19.21 + j 3.52

V

Zs

ZL

Vs

10 m

Figure 3.12

Zs

Vs

Circuit arrangement for Example 3.11. ZTh

VA

Zin

VTh

ZL

Figure 3.13 Equivalent circuits at the two ends of the transmission line shown in Figure 3.12.

78

TRANSMISSION LINES

or 1952.78 0.181 rad ◦ = 28.1788 V 0.1302 rad = 28.1788 V 7.46 69.2995 0.0508 rad

VA =

For determining voltage at the load, a Th´evenin equivalent circuit can be used as follows: ZTh = Z0

Zs + j Z0 tan βl = Z0 Z0 + j Zs tan βl

(∵Zs = Z0 = 50 )

and VTh = (Vin + Vref )at

o.c.

= (2 × Vin )at

o.c.

= 100 V −8.168 rad

Therefore, VL =

VTh 100 −8.168 rad ZL = × (100 + j 50) V ZTh + ZL 50 + 100 + j 50

or ◦

◦

VL = 70.71 V −8.0261 rad = 70.71 V −459.86 = 70.71 V −99.86 Alternatively, V (z) = Vin e−j βz + Vref ej βz = Vin (e−j βz + ej βz )

where Vin is incident voltage at z = 0 and is the input reflection coefficient. ◦

Vin = 50 V 0 and =

19.21 + j 3.52 − 50 = 0.4472 2.9769 rad 19.21 + j 3.52 + 50

Hence, VL = V (z = 10 m) = 50(e−j 8.168 + 0.4472ej (2.9769+8.168) ) = 50 × 1.4142 V −1.743 rad or VL = 70.7117 V −99.8664◦ Example 3.12 Two identical signal generators are connected in parallel through a quarter-wavelength-long lossless 50- transmission line (Figure 3.14). Each of these generators has an open-circuit voltage of 12 0◦ V and a source impedance

79

SENDING-END IMPEDANCE

Z0 = 50 Ω

Z0 = 50 Ω

50 Ω

A

50 Ω

12 V∠0°

B

12 V∠0°

λ/4

Figure 3.14

10 Ω

λ/4

Circuit arrangement for Example 3.12.

of 50 . It drives a 10- load through another quarter-wavelength-long similar transmission line. Determine the power dissipated by the load. SOLUTION The signal generator connected at the left (source A) and the quarter-wavelength-long line connecting it can be replaced by a Th´evenin equivalent voltage source of 12 V −90◦ with its internal resistance at 50 . This voltage source can be combined with the 12 0◦ V (source B) already there, which also has an internal resistance√of 50 . The resulting source will have a Th´evenin voltage of 6 − j 6 V = 6 2 V −45◦ and an internal resistance of 25 (i.e., two 50- resistances connected in parallel). The load will transform to 50 × 50/10 = 250 at the location of source B. Therefore, a simplified circuit can be drawn as in Figure 3.15. The voltage across 250 will be equal to 250 ×

√ 6 − j6 ◦ = 60 2/11 −45 V 275 V

Thus, the dissipated power, Pd , can be calculated as follows: Pd =

602 × 2 = 0.119008 W = 119.008 mW 112 × 2 × 250

25 Ω

6√2 V∠− 45°

Figure 3.15

250 Ω

Simplification of the circuit shown in Figure 3.14.

80

TRANSMISSION LINES

3.3 STANDING WAVE AND STANDING WAVE RATIO Consider a lossless transmission line that is terminated by a load impedance ZL , as shown in Figure 3.16. Incident and reflected voltage phasors at its input (i.e., at z = 0) are assumed to be Vin and Vref , respectively. Therefore, total voltage V (z) can be expressed as follows: V (z) = Vin e−j βz + Vref e+j βz Alternatively, V (x) can be written as V (x) = V+ e+j βx + V− e−j βx = V+ (e+j βx + e−j βx ) or

V (x) = V+ [e+j βx + ρe−j (βx−φ) ]

where = ρej φ =

(3.3.1)

V− V+

V+ and V− represent incident and reflected wave voltage phasors, respectively, at the load point (i.e., at x = 0). Let us first consider two extreme conditions at the load. In one case, the transmission line has an open circuit, whereas in the other, it has a short-circuit termination. Therefore, the magnitude of the reflection coefficient, ρ, is unity in both cases. However, the phase angle, φ, is zero for the former case and π for the latter. Phasor diagrams for these two cases are depicted in Figure 3.17(a) and (b), respectively. As distance x from the load increases, the phasor ej βx rotates counterclockwise because of the increase in its phase angle βx. On the other hand, the phasor e−j βx rotates clockwise by the same amount. Therefore, the phase angle of the resulting voltage, V (x), remains constant with space coordinate x while its magnitude varies sinusoidally between ±2V+ . Since the phase angle of the resulting signal V (x) does not change with distance, it does not represent

V+ V−

Vin Vref

z

x

Figure 3.16 Lossless transmission line with termination.

81

STANDING WAVE AND STANDING WAVE RATIO

e jβx βx V(x)

V(x) −βx

e−j(βx − π) π − βx

e−jβx

e jβx βx

(b)

(a)

Figure 3.17 Phasor diagrams of line voltage with (a) open-circuit and (b) short-circuit termination.

a propagating wave. Note that there are two waves propagating on this line in opposite directions. However, the resulting signal represents a standing wave. When the terminating load is of an arbitrary value that is different from the characteristic impedance of the line, there are still two waves propagating in opposite directions. The interference pattern of these two signals is stationary with time. Assuming that V+ is unity, a phasor diagram for this case is drawn as shown in Figure 3.18. The magnitude of the resulting signal, V (x), can be determined using the law of parallelograms as follows: |V (x)| = |V+ |[1 + ρ2 + 2ρ cos(2βx − φ)]1/2

ejβx βx

V(x)

θ

−(βx − φ) ρe−j(βx − φ)

(2βx − φ)

Figure 3.18 Phasor diagram of line voltage with arbitrary termination.

(3.3.2)

82

TRANSMISSION LINES

The reflection coefficient, (x), can be expressed as (x) =

V− e−j βx = ρe−j (2βx−φ) V+ ej βx

(3.3.3)

Hence, the magnitude of this standing wave changes with location on the transmission line. Since x appears only in the argument of cosine function, the voltage magnitude has an extreme value whenever this argument is an integer multiple of π. It has a maximum whenever the reflected wave is in phase with the incident signal, which requires that the following condition be satisfied: 2βx − φ = ±2nπ,

n = 0, 1, 2, . . .

(3.3.4)

On the other hand, V (x) has a minimum value where the reflected and incident signals are out of phase. Hence, x satisfies the following condition at a minimum of the interference pattern: 2βx − φ = ±(2m + 1)π,

m = 0, 1, 2, . . .

(3.3.5)

Further, these extreme values of the standing waves are |V (x)|max = |V+ |(1 + ρ)

(3.3.6)

|V (x)|min = |V+ |(1 − ρ)

(3.3.7)

and

The ratio of maximum to minimum values of voltage V (x) is called the voltage standing wave ratio (VSWR). Therefore, VSWR = S =

|V (x)|max 1+ρ = |V (x)|min 1−ρ

(3.3.8)

Since 0 ≤ ρ ≤ 1 for a passive load, the minimum value of the VSWR will be unity (for a matched load) while its maximum value can be infinity (for total reflection, with a short circuit or an open circuit as the load). Assume that there is a voltage minimum at x1 from the load, and if one keeps moving toward the source, the next minimum occurs at x2 . In other words, there are two consecutive minimums at x1 and x2 with x2 > x1 . Hence, 2(βx1 − φ) = (2m1 + 1)π and 2(βx2 − φ) = [2(m1 + 1) + 1]π Subtracting the former equation from the latter, one finds that 2β(x2 − x1 ) = 2π ⇒ x2 − x1 =

λ 2

where x2 − x1 is the separation between the two consecutive minimums.

83

STANDING WAVE AND STANDING WAVE RATIO

Similarly, it can be proved that two consecutive maximums are a half-wavelength apart and also that separation between the consecutive maximum and minimum is a quarter-wavelength. This information can be used to measure the wavelength of a propagating signal. In practice, the location of a minimum is preferred over that of a maximum. This is because minimums are sharper in comparison with maximums, as illustrated in Figure 3.19. Further, a short (or open) circuit must be used as the load for the best measurement accuracy. Measurement of Impedance Impedance of a one-port microwave device can be determined from measurement on the standing wave at its input. Those required parameters are the VSWR and the location of the first minimum (or maximum) from the load. A slotted line that is equipped with a detector probe is connected before the load to facilitate measurement. Since the output of a detector is proportional to power, the square root of the ratio is taken to find the VSWR. Since it may not be possible in most cases to probe up to the input terminals of the load, the location of the first minimum is determined as follows. An arbitrary minimum is located on the slotted line with unknown load. The load is then replaced by a short circuit. As a result, there is a shift in minimum, as shown in Figure 3.19. The shift of the original minimum away from the generator is equal to the location of the first minimum from the load. Since the reflected voltage is out of phase with that of the incident signal at the minimum of the standing wave pattern, a relation between the reflection coefficient, 1 = −ρ, and the impedance Z1 at this point may be written as 1 = −ρ =

2.0

Z1 − 1 Z1 + 1

With unknown load

With short circuit

Magnitude of V(x)

1.5

1.0

0.5

0.0 0.0

1.0

2.0

3.0

4.0

x

Figure 3.19

Standing wave pattern on a lossless transmission line.

5.0

84

TRANSMISSION LINES

where Z 1 is the normalized impedance at the location of the first minimum of the voltage standing wave. Since S= and therefore, −

1+ρ 1−ρ

and ρ =

S−1 S+1

1 Z1 − 1 S−1 = ⇒ Z1 = S+1 S Z1 + 1

This normalized impedance is equal to the input impedance of a line that is terminated by load ZL and has a length of d1 . Hence, Z1 =

1 1 − j S tan βd1 Z L + j tan βd1 ⇒ ZL = = S S − j tan βd1 1 + j Z L tan βd1

Similarly, the reflected voltage is in phase with the incident signal at the maximum of the standing wave. Assume that the first maximum is located at a distance d2 from the unknown load and that the impedance at this point is Z2 . One can write S−1 Z2 − 1 2 = ρ = = ⇒ Z2 = S S+1 Z2 + 1 and ZL =

Z 2 − j tan βd2 1 − j Z 2 tan βd2

Since the maximum and minimum are measured on a lossless line that feeds the unknown load, magnitude of the reflection coefficient ρ does not change at different points on the line. Example 3.13 A load impedance of 73 − j 42.5 terminates a transmission line of characteristic impedance 50 + j 0.01 . Determine its reflection coefficient and the voltage standing wave ratio. SOLUTION = =

ZL − Z0 73 − j 42.5 − (50 + j 0.01) = ZL + Z0 73 − j 42.5 + (50 + j 0.01) 23 − j 42.51 48.3332 −1.0749 rad = 123 − j 42.49 130.1322 −0.3326 rad

◦

= 0.3714 −0.7423 rad = 0.3714 −42.53 and VSWR =

1+ρ 1 + 0.3714 = = 2.1817 1−ρ 1 − 0.3714

SMITH CHART

85

Note that this line is lossy, and therefore the reflection coefficient and VSWR will change with distance from the load. Example 3.14 A 100- transmission line is terminated by the load ZL . Measurements indicate that it has a VSWR of 2.4 and the standing wave minimums are 100 cm apart. The scale reading at one of these minimums is found to be 275 cm. When the load is replaced by a short circuit, the minimum moves away from the generator to a point where the scale shows 235 cm. Find the signal wavelength and the load impedance. SOLUTION Since the minimums are 100 cm apart, the signal wavelength λ = 2 × 100 = 200 cm = 2 m, β = π rad/m, and d1 = 275 − 235 = 40 cm. Hence, ZL =

1 − j 2.4 tan(0.4π) = 1.65 − j 0.9618 2.4 − j tan(0.4π)

or ZL = 165 − j 96.18 3.4 SMITH CHART Normalized impedance at a point in the circuit is related to its reflection coefficient as 1 + r + j i 1+ = Z = R + jX = (3.4.1) 1− 1 − r − j i where r and i represent real and imaginary parts of the reflection coefficient, while R and X are real and imaginary parts of the normalized impedance, respectively. This complex equation can be split into two, after equating real and imaginary components on its two sides. Hence, 2 2 1 R 2 + i = (3.4.2) r − 1+R 1+R and

2 1 2 1 = (r − 1) + i − X X 2

(3.4.3)

These two equations represent a family of circles on the complex -plane. The circle in (3.4.2) has its center at (R/(1 + R), 0) and a radius of 1/(1 + R). For R = 0 it is centered at the origin with unity radius. As R increases, the center of the constant-resistance circle moves on a positive real axis and its radius decreases. When R = ∞, the radius reduces to zero and the center of the circle moves to (1,0). These plots are shown in Figure 3.20. Note that for passive impedance, 0 ≤ R ≤ ∞ and −∞ ≤ X ≤ +∞. Similarly, (3.4.3) represents a circle that is centered at (1, 1/X) with a radius of 1/X. For X = 0, its center lies at (1,∞) with an infinite radius. Hence, it is a

86

TRANSMISSION LINES

straight line along the r -axis. As X increases on the positive side (i.e., 0 ≤ X ≤ ∞), the center of the circle moves toward point (1,0) along a vertical line defined by r = 1, and its radius becomes smaller and smaller in size. For X = ∞, it becomes a point that is located at (1,0). Similar characteristics are observed for 0 ≥ X ≥ −∞. As shown in Figure 3.20, a graphical representation of these two equations for all possible normalized resistance and reactance values is known as the Smith chart. Thus, a normalized impedance point on the Smith chart represents the corresponding reflection coefficient in polar coordinates on the complex -plane. According to (3.2.7), the magnitude of the reflection coefficient on a lossless transmission line remains constant as ρL while its phase angle decreases as −2βl. Hence, it represents a circle of radius ρL . As one moves away from the load (i.e., toward the generator), the reflection coefficient point moves clockwise on this circle. Since the reflection coefficient repeats periodically at every half-wavelength, the circumference of the circle is equal to λ/2. For a given reflection coefficient, normalized impedance may be found using the impedance scale of the Smith chart. Further, R in (3.4.1) is equal to the VSWR for r > 0 and i = 0. On the other hand, it equals the inverse of VSWR for r < 0 and i = 0. Hence, R values on the positive r -axis also represent the VSWR.

Γi 90° (1,1)

Constant R or G circles

1.0 0.5

2.0

0.2

0°

180° 0

0.1

0.5

1.0

2.0

5.0

Constant X or B arcs 0.2 2.0 0.5 1.0 270° 1

Figure 3.20

Smith chart.

Γr

SMITH CHART

87

Since admittance represents the inverse of impedance, we can write Y =

1 Z

=

1− 1+

Hence, a similar analysis can be performed with the normalized admittance instead of impedance. It results in the same kind of chart except that the normalized conductance circles replace normalized resistance circles, while the normalized susceptance arcs replace normalized reactance. Normalized resistance (or conductance) of each circle is indicated on the r -axis of the Smith chart. Normalized positive reactance (or susceptance) arcs are shown on the upper half, while negative reactance (or susceptance) arcs are seen in the lower half. The Smith chart in conjunction with equation (3.2.7) facilitates the analysis and design of transmission line circuits. Example 3.15 A load impedance of 50 + j 100 terminates a lossless, quarterwavelength-long transmission line. If characteristic impedance of the line is 50 , find the impedance at its input end, the load reflection coefficient, and the VSWR on this transmission line. SOLUTION This problem can be solved using equations (3.2.6), (3.2.7), and (3.3.8) or the Smith chart. Let us try it both ways. 2π λ π ◦ × = = 90 λ 4 2 ZL + j Z0 tan βl (50 + j 100) + j 50 tan(90◦ ) Zin = Z0 = 50 Z0 + j ZL tan βl 50 + j (50 + j 100) tan(90◦ ) βl =

or Zin = 50

j 50 2500 = = 10 − j 20 j (50 + j 100) 50 + j 100

50 + j 100 − 50 j 100 100 90◦ ZL − Z0 = = = √ ZL + Z0 50 + j 100 + 50 100 + j 100 100 × 2 45◦ ◦ = 0.7071 45

=

VSWR =

1 + || 1 + 0.7071 1.7071 = = = 5.8283 1 − || 1 − 0.7071 0.2929

To solve this problem graphically using the Smith chart, normalized load impedance is determined as follows: ZL =

ZL 50 + j 100 = = 1 + j2 Z0 50

This point is located on the Smith chart as shown in Figure 3.21. A circle that passes through 1 + j 2 is then drawn with point 1 + j 0 as its center. Since the

88

1

0.9

0.7

0.15 0.35

2

4

0 0.2 30 0.

0.

THS TOWARD GENER E NG A VEL-- WAVELENGTHS TOWARD TOR A LO ---> - W AV D- W - WAV 0.0 0.0 5 0.0 -WA 0 is a perfect dielectric with εr = 2.25, and the region x < 0 is a free space. At the interface, subscript 1 denotes field components on the +x side of the boundary and subscript 2 those on the −x side. If D1 = 2 xˆ + 2yC/m ˆ , find D2 , E1 , and E2 . SOLUTION The conditions stated in the problem are illustrated in Figure 4.10. The interface is x = 0 plane, and therefore the x component, will be normal, with

Medium 2 Free space

y

Medium 1 εr = 2.25

x

z

Figure 4.10

Conditions for Example 4.8.

122

ELECTROMAGNETIC FIELDS AND WAVES

the y component tangential to this plane. D1 = ε1 E1 → E1 =

D1 1 2 = xˆ + yˆ ε1 2.25ε0 2.25ε0

V/m

From the boundary conditions, the x component of electric flux density and y component of electric field intensity in medium 2 will be same as in medium 1. Hence, 2 D2x = 1 and E2y = 2.25ε0 Therefore, D2 = xˆ +

2 yˆ 2.25

C/m2

and E2 =

D2 1 2 = xˆ + yˆ ε0 ε0 2.25ε0

V/m

Example 4.9 A sphere of 2 m radius is made of a perfect dielectric material (medium 1). It is surrounded by free space (medium 2). Electric field intensities in the two media are given as follows: ˆ E1 = E01 (cos θˆr − sin θθ)

and E2 = E02

1+

8 r3

r ≤ 2m

4 cos θˆr − 1 − 3 sin θθˆ r

r ≥ 2m

Find the permittivity of the spherical medium. SOLUTION As indicated in Figure 4.11, the r components of the electric fields are normal and the θ components are tangential at the interface. Therefore, 4 sin θE02 → E02 = 2E01 − sin θE01 = − 1 − 8 z

θ r y

x

Figure 4.11

φ

Conditions for Example 4.9.

UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE

123

and ε1 E01 cos θ = ε0 E02 ž 2 cos θ → ε1 = 2ε0

E02 = 4ε0 E01

F/m

4.4 UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE Consider a uniform plane electromagnetic wave propagating in medium 1 as illustrated in Figure 4.12. The electric field Ei of this wave is y directed, and its magnetic field Hi is in the −x direction. A wave propagating in the +z direction is incident normally on the interface of medium 2 with medium 1. The electrical characteristics of the two media are as indicated in Figure 4.12. Since the wave experiences a change in electrical characteristics in its path, a part of the incident signal is reflected back, while the remaining signal is transmitted into medium 2. Thus, there are two waves in medium 1 propagating in opposite directions, and one in medium 2 that continues along +z. The electromagnetic fields associated with the incident wave can be expressed as follows: Ei = yE ˆ 0 e−j k1 z

(4.4.1)

and Hi = −xH ˆ 0 e−j k1 z = −xˆ

E0 −j k1 z e η1

(4.4.2)

x

Medium 1 σ1,ε1,µ1

Medium 2 σ2,ε2,µ2 Et

Ei Hr

Hi

Ht

Er z y

Figure 4.12 Geometry of a uniform plane wave incident normally on the interface of two media.

124

ELECTROMAGNETIC FIELDS AND WAVES

where E0 and H0 are arbitrary constants, k1 the complex wave number, and η1 the complex intrinsic impedance of medium 1. Similarly, electromagnetic fields associated with the reflected wave (Er and r H ) as well as the transmitted wave (Et and Ht ) can be expressed as follows: Er = yRE ˆ 0 e+j k1 z −j k1 z Hr = xRH ˆ = xR ˆ 0e

(4.4.3) E0 +j k1 z e η1

(4.4.4)

Et = yTE ˆ 0 e−j k2 z

(4.4.5)

and −j k1 z Ht = −xTH ˆ = −xT ˆ 0e

E0 −j k2 z e η2

(4.4.6)

where R and T are complex reflection and transmission coefficients, respectively, k2 the complex wave number in medium 2, and η2 the complex intrinsic impedance of medium 2. Now, applying the boundary conditions (tangential components of the fields are continuous across the boundary) at z = 0 gives E0 + RE 0 = TE 0 → 1 + R = T

(4.4.7)

and −

E0 E0 η1 E0 +R = −T →1−R = T η1 η1 η1 η2

(4.4.8)

Equations (4.4.7) and (4.4.8) can be solved for R and T as follows: T =

2η2 η1 + η2

(4.4.9)

R=

η2 − η1 η1 + η2

(4.4.10)

and

Therefore, the total electric field at any point in medium 1 is found to be ˆ 0 e−j k1 z + yRE ˆ 0 e+j k1 z E1 = yE |E1 |2 = |E0 |2 [e−2α1 z + ρ2 e2α1 z + 2ρ cos(2β1 z + θ)]

(4.4.11)

where R = ρej θ

(4.4.12)

k1 = β1 − j α1

(4.4.13)

UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE

125

β1 is the phase constant in rad/s and α1 is the attenuation constant in Np/m for medium 1. It is to be noted that the propagation characteristics of this wave are very much similar to that of a wave propagating on the transmission line, as discussed in Chapter 3. Example 4.10 The electric field intensity of a uniform plane wave propagating in free space is given as Ei (z, t) = yˆ ž 4.8 cos(109 t − k0 z)

V/m

If this wave impinges normally on a lossless medium (µr = 1 and εr = 2.25) in the z ≥ 0 region, find k0 , k2 , R, T , the VSWR, and the average power transmitted into the medium. SOLUTION The wave numbers and intrinsic impedance of the wave in the two media are found as follows: √ k0 = ω µ0 ε0 = 109 4π × 10−7 × 8.854 × 10−12 = 3.3356 rad/m √ √ k2 = ω µ0 ε0 εr = k0 2.25 = 5.0034 rad/m

µ0 4π × 10−7 η1 = = = 376.7343 ε0 8.854 × 10−12 and η2 =

µ0 = ε0 εr

4π × 10−7 = 251.1562 8.854 × 10−12 × 2.25

Now R and T can be found via (4.4.9) and (4.4.10) as η2 − η1 ≈ −0.2 η1 + η2

R= and T =

2η2 ≈ 0.8 η1 + η2

The VSWR in medium 1 is found to be VSWR =

1+ρ 1 + 0.2 = = 1.5 1−ρ 1 − 0.2

The electric field intensity in medium 2 is Et (z, t) = yˆ ž 3.84 cos(109 t − k2 z)

V/m

126

ELECTROMAGNETIC FIELDS AND WAVES

Therefore, the time-averaged power transmitted into medium 2 may be found as follows: 1 1 |Et |2 1 3.842 = Pav = Re(E × H∗ ) = = 0.0294 W/m2 2 2 η2 2 251.1562 4.5 MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS Equation (4.1.9) implies that the magnetic flux does not start from or terminate at a magnetic charge the way that electric flux does with electric charge. This is because there is no real magnetic charge (or magnetic monopole). In fact, the smallest unit of the magnetic source is a magnetic dipole that is actually an infinitesimal current loop. Consider a general case where the source consists of two types of currents: a linear current that flows or oscillates in the linear direction, and an infinitesimal circulatory current that flows or oscillates around an infinitesimal loop. For this case it is rather hard to describe mathematically the electric current density J that includes both parts. Mathematically, it is sometimes advantageous to recognize its linear part as electric current Je and imagine its infinitesimal circulatory current as an equivalent magnetic dipole source. If the circulatory current varies in magnitude and direction or oscillates around the loop with time, the equivalent magnetic dipole strength also oscillates with time. When the magnetic dipole oscillates with time, it can be considered as an imaginary magnetic current flowing up and down. This is completely analogous to a time-varying electric dipole. Therefore, a time-varying circulatory current can be represented by an equivalent magnetic current and charges, Jm and ρm , respectively. Imagine now that the magnetic charge ρm produces a magnetic flux, analogous to the electric charge ρe producing the electric flux. Then equation (4.1.9) should be modified to ∇ ž B = ρm If (4.1.9) is modified, (4.1.6) also needs modification because divergence of the curl of a vector is always zero, which is not true because ρm is a function of time: ∂ ∂B ∂ ∇ ž (∇ × E) = −∇ ž = − (∇ ž B) = − ρm = 0 ∂t ∂t ∂t If it is assumed that the usual continuity equation should hold between ρm and Jm , ∇ ž Jm = −

∂ρm ∂t

Hence, equation (4.1.6) should be modified to ∇ × E = −J m −

∂B ∂t

MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS

127

Therefore, the modified Maxwell’s equations for the time-harmonic field and the two continuity equations can be expressed as: ∇ × E(r) = −Jm (r) − j ωB(r)

(4.5.1)

∇ × H(r) = Je (r) + j ω D(r)

(4.5.2)

∇ ž D(r) = ρe (r)

(4.5.3)

∇ ž B(r) = ρm (r)

(4.5.4)

∇ ž Je (r) = −j ωρe (r)

(4.5.5)

∇ ž Jm (r) = −j ωρm (r)

(4.5.6)

and

Magnetic Vector and Electric Scalar Potentials Potential functions are introduced to transform Maxwell’s equations mathematically and so facilitate solutions to electromagnetic problems. Consider that there are only electric sources (current and charge) present in a volume under consideration (i.e., Je = 0, ρe = 0, Jm = 0, and ρm = 0). Therefore, Maxwell’s equations for time-harmonic fields and the equation of continuity can be written as follows: ∇ × E = −j ωB

(4.5.7)

∇ × H = Je + j ωD

(4.5.8)

∇ ž D = ρe

(4.5.9)

∇žB = 0

(4.5.10)

∇ ž Je = −j ωρe

(4.5.11)

and

As defined earlier, the constitutive relations are B = µH

(4.5.12)

D = εE

(4.5.13)

and

Since divergence of the curl of a vector is always zero, the magnetic flux density in (4.5.10) may be assumed to be the curl of vector A. Hence, ∇žB = 0 ⇒ B = ∇ × A

(4.5.14)

128

ELECTROMAGNETIC FIELDS AND WAVES

Substitution of B from (4.5.14) into (4.5.7), and recognizing the fact that the curl of the gradient of a scalar is always zero gives ∇ × E = −j ω ∇ × A ⇒ ∇×(E + j ωA) = 0 ⇒ E + j ωA = −∇φe Therefore, E = −∇φe − j ωA

(4.5.15)

where A is called the magnetic vector potential and φe is known as the electric scalar potential. Using (4.5.12), (4.5.13), and (4.5.15), (4.5.8) may be written as ∇×

∇×A = Je + j ωεE = Je + j ωε(−∇φe − j ωA) µ

If µ is not changing with space coordinates, this relation simplifies further as follows: ∇ × ∇ × A = ∇(∇ ž A) − ∇ 2 A = µJe − j ωεµ∇φe + ω2 µεA or ∇ 2 A + ω2 µεA = −µJe + j ωεµ∇φe + ∇(∇ ž A) or ∇ 2 A + ω2 µεA = −µJe + ∇(∇ ž A + j ωεµφe )

(4.5.16)

Similarly for ε not changing with space coordinates, (4.5.9), (4.5.13), and (4.5.15) give ∇ ž (−∇φe − j ωA) =

ρe ρe → ∇ 2 φe + j ω(∇ ž A) = − ε ε

(4.5.17)

Equations (4.5.16) and (4.5.17) represent a pair of coupled partial differential equations for φ and A. These may be uncoupled via Helmholtz’s theorem, which states that a vector is specified completely by its curl and divergence. Since only the curl of A is defined via (4.5.14), we are at liberty to specify its divergence. The Lorentz condition may be used to define it as ∇ ž A = −j ωεµφe

(4.5.18)

Thus, (4.5.16) and (4.5.17) simplify to ∇ 2 A + k 2 A = −µJ

(4.5.19)

ρe ε

(4.5.20)

and ∇ 2 φe + k 2 φe = −

129

MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS

where k 2 = ω2 µε and k is the wave number. It should be noted at this point that the continuity equation is implied by the Lorentz condition.

Electric Vector and Magnetic Scalar Potentials Electric vector and magnetic scalar potentials are introduced when the existence of only magnetic current and charge is assumed in the region (i.e, ρe = 0 and Je = 0). Therefore, Maxwell’s equations and the equation of continuity for this case are found to be ∇ × E = −Jm − j ωB

(4.5.21)

∇ × H = j ωD

(4.5.22)

∇žD = 0

(4.5.23)

∇ ž B = ρm

(4.5.24)

and ∇ ž Jm = −j ωρm

(4.5.25)

As before, D = εE and B = µH. Because of (4.5.23), it is assumed that D = −∇ × F

(4.5.26)

where F is called the electric vector potential. After substituting (4.5.26) into (4.5.22) and noting that curl of the gradient of a scalar is always zero, we get ∇ × H = −j ω∇ × F → ∇×(H + j ωF) = 0 → H + j ωF = −∇φm Therefore, H = −∇φm − j ωF

(4.5.27)

where φm is called the magnetic scalar potential. Now, from (4.5.21), (4.5.26), and (4.5.27),

∇×F ∇× − ε

= −Jm − j ωµ(−∇φm − j ωF)

or −∇ × ∇ × F = −εJm + j ωµε∇φm − ω2 µεF or −[∇(∇ ž F) − ∇ 2 F] + ω2 µεF = −εJm + j ωµε∇φm

(4.5.28)

130

ELECTROMAGNETIC FIELDS AND WAVES

As in the preceding case, if the Lorentz condition ∇ ž F = −j ωµεφm

(4.5.29)

∇ 2 F + ω2 µεF = −εJm

(4.5.30)

is introduced, then Further, (4.5.24), (4.5.27), and (4.5.29) give ∇ ž [µ(−∇φm − j ωF)] = ρm ⇒ −∇ 2 φm − j ω(−j ωµεφm ) = or ∇ 2 φm + ω2 µεφm = −

ρm µ

ρm µ (4.5.31)

An electromagnetic field problem can be divided into these two cases, solved for each case, and the individual solutions can be superimposed to find the complete solution. Thus, the electromagnetic fields may be found easily after determining the vector potentials via (4.5.19) and (4.5.30) and the corresponding Lorentz conditions. This procedure is used in the following section to further formulate general solution techniques for source-free cases that are specialized subsequently for a few waveguides.

4.6 CONSTRUCTION OF SOLUTIONS The analysis presented in Section 4.5 is specialized here for a source-free region with rectangular as well as cylindrical coordinates. It is further divided into two cases; the electric vector potential is assumed to be zero in one, and the magnetic vector potential, zero in the other. Rectangular Coordinate System 1. Assume that A = zˆ Az and F = 0; then (4.5.14), (4.5.15), and (4.5.18) give

1 1 ∂Az ∂Az H = ∇×ˆzAz = xˆ + yˆ − + zˆ 0 (4.6.1) µ µ ∂y ∂x and E = −j ωˆzAz +

1 ∂Az ∇ j ωµε ∂z

(4.6.2)

For this case, (4.5.19) simplifies to ∇ 2 Az + k 2 Az = 0 →

∂ 2 Az ∂ 2 Az ∂ 2 Az + + + k 2 Az = 0 ∂x 2 ∂y 2 ∂z2

(4.6.3)

CONSTRUCTION OF SOLUTIONS

131

After (4.6.3) is solved for a given problem, field components are found from (4.6.1) and (4.6.2) as follows: Hx =

1 ∂Az µ ∂y

Hy = −

(4.6.4)

1 ∂Az µ ∂x

(4.6.5)

Hz = 0

(4.6.6) 2

Ex =

1 ∂ Az j ωµε ∂x∂z

(4.6.7)

Ey =

1 ∂ 2 Az j ωµε ∂y∂z

(4.6.8)

Ez =

1 j ωµε

and

∂2 2 + k Az ∂z2

(4.6.9)

Since the magnetic fields are transverse to the z-axis (Hz = 0), these fields are known as TMz (transverse magnetic to z ) mode fields. 2. Assume that F = zˆ Fz and A = 0; then (4.5.26) and (4.5.27) give

∂Fz 1 1 ∂Fz + yˆ − E = − ∇×ˆzFz = − xˆ + zˆ 0 ε ε ∂y ∂x

(4.6.10)

and H = −j ωˆzFz +

1 ∂Fz ∇ j ωµε ∂z

(4.6.11)

For this case, (4.5.30) simplifies to ∇ 2 Fz + k 2 Fz = 0 →

∂ 2 Fz ∂ 2 Fz ∂ 2 Fz + + + k 2 Fz = 0 ∂x 2 ∂y 2 ∂z2

(4.6.12)

After (4.6.12) is solved for a given problem, field components are found from (4.6.10) and (4.6.11) as follows: Ex = −

1 ∂Fz ε ∂y

(4.6.13)

1 ∂Fz ε ∂x Ez = 0

Ey =

Hx =

(4.6.14) (4.6.15) 2

1 ∂ Fz j ωµε ∂x ∂z

(4.6.16)

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ELECTROMAGNETIC FIELDS AND WAVES

Hy =

1 ∂ 2 Fz j ωµε ∂y∂z

Hz =

1 j ωµε

and

∂2 2 + k Fz ∂z2

(4.6.17)

(4.6.18)

Since the electric fields are transverse to the z-axis in this case, these fields are known as TE z (transverse electric to z ) mode fields. Cylindrical Coordinate System 1. Assume that A = zˆ Az and F = 0; then from (4.5.14), (4.5.15), and (4.5.18), we get

∂Az 1 1 1 ∂Az ˆ +φ − H = ∇×ˆzAz = ρˆ + zˆ 0 (4.6.19) µ µ ρ ∂φ ∂ρ and E = −j ωˆzAz +

∂Az 1 ∇ j ωµε ∂z

(4.6.20)

For this case, (4.5.19) in cylindrical coordinates may be written as follows: 1 ∂ ∂ 2 Az ∂Az 1 ∂ 2 Az 2 2 ∇ Az + k Az = 0 → + + k 2 Az = 0 (4.6.21) ρ + 2 ρ ∂ρ ∂ρ ρ ∂φ2 ∂zz Once solutions to (4.6.21) are found, electromagnetic fields may be determined from (4.6.19) and (4.6.20) as follows: Hρ =

1 ∂Az µρ ∂φ

(4.6.22)

1 ∂Az µ ∂ρ

(4.6.23)

Hφ = − Hz = 0

and

(4.6.24) 2

Eρ =

1 ∂ Az j ωµε ∂ρ∂z

(4.6.25)

Eφ =

1 ∂ 2 Az j ωµερ ∂φ∂z

(4.6.26)

1 Ez = j ωµε

∂2 2 + k Az ∂z2

(4.6.27)

METALLIC PARALLEL-PLATE WAVEGUIDE

133

This is the TMz mode (magnetic field transverse to the z-axis) in a cylindrical coordinate system. 2. For F = zˆ Fz and A = 0, (4.5.26) and (4.5.27) in cylindrical coordinates give

1 1 1 ∂Fz ∂Fz E = − ∇×ˆzFz = − ρˆ + φˆ − + zˆ 0 (4.6.28) ε ε ρ ∂φ ∂ρ and

1 ∂Fz H = −j ωˆzFz + ∇ j ωµε ∂z

(4.6.29)

Further, (4.5.30) reduces to 1 ∂ ∂ 2 Fz ∂Fz 1 ∂ 2 Fz + + k 2 Fz = 0 (4.6.30) ρ + 2 ∇ Fz + k Fz = 0 → ρ ∂ρ ∂ρ ρ ∂φ2 ∂z2 2

2

After (4.6.30) is solved, the field components are determined from (4.6.28) and (4.6.29) as follows: Eρ = −

1 ∂Fz ερ ∂φ

1 ∂Fz ε ∂ρ Ez = 0

Eφ =

(4.6.31) (4.6.32) (4.6.33)

2

Hρ =

1 ∂ Fz j ωµε ∂ρ∂z

(4.6.34)

Hφ =

1 ∂ 2 Fz j ωµερ ∂φ ∂z

(4.6.35)

Hz =

1 j ωµε

and

∂2 2 + k Fz ∂z2

(4.6.36)

This is the TEz mode (electric field transverse to the z-axis) in a cylindrical coordinate system. These formulations are used in the following sections to analyze propagation of electromagnetic signals through the waveguide. 4.7 METALLIC PARALLEL-PLATE WAVEGUIDE Consider a parallel-plate waveguide that consists of two perfectly conductive plates extending to infinity on the y –z plane, as illustrated in Figure 4.13. Separation between the two plates is assumed to be a. There is an electromagnetic

134

ELECTROMAGNETIC FIELDS AND WAVES z

a x

y

Figure 4.13 Metallic parallel-plate waveguide.

signal that propagates along the z-axis with a complex propagation constant γ. Therefore, the field variation along the z-axis is assumed to be e−γz , whereas it stays constant along the y-axis (i.e., ∂/∂y → 0). For analyzing TM modes between the two plates, partial differential equation (4.6.3) simplifies to the following ordinary differential equation: d 2 Az (x) + (k 2 + γ2 )Az (x) = 0 dx 2

(4.7.1)

A solution to this equation can be found easily as follows: Az (x) = c1 cos kx x + c2 sin kx x

(4.7.2)

where c1 and c2 are integration constants to be evaluated from the boundary conditions, and kx2 = k 2 + γ2 (4.7.3) Therefore,

Az (x, y, z) = (c1 cos kx x + c2 sin kx x)e−γz

(4.7.4)

Since the boundary conditions require that the tangential electric fields must be zero on the conducting surfaces at x = 0 and at x = a, Ez must be zero on these surfaces. It is to be noted that Ey is zero everywhere because Az is independent of y. For a nontrivial solution, Ez | x=0 = 0 → Az (x, y, z)| x=0 = 0 x=a

x=a

(4.7.5)

Equation (4.7.5) is satisfied only if c1 = 0

(4.7.6)

135

METALLIC PARALLEL-PLATE WAVEGUIDE

and kx =

mπ , a

Therefore, Az = c2 sin

m = 0, 1, 2, . . .

(4.7.7)

mπ x e−γz a

(4.7.8)

Further, the propagation constant can be found from (4.7.3) as follows: mπ 2 γ= − k2 (4.7.9) a Therefore, the signal will propagate without attenuation if k > mπ/a, and it will attenuate for k < mπ/a. The cutoff occurs at k = mπ/a. Once Az is determined, the field components may be found from (4.6.4)–(4.6.9) as follows: Hx = 0

(4.7.10)

mπ c2 mπ Hy = − cos x e−γz µ a a

(4.7.11)

Hz = 0

(4.7.12)

mπ γ mπ cos x e−γz Ex = −c2 j ωµε a a

(4.7.13)

Ey = 0

(4.7.14)

and Ez =

mπ c2 mπ 2 x e−γz sin j ωµε a a

(4.7.15)

These results are summarized in Table 4.1 assuming that −c2 (kc /µ) = E0 . A similar procedure may be used to determine the characteristics of TE mode fields propagating through this waveguide. Final results are summarized in Table 4.1. Example 4.11 A metallic parallel-plate waveguide is air-filled, and the separation between its plates is 4.5 cm. Investigate the characteristics of a 12-GHz signal propagating in TMm0 modes. SOLUTION kc =

mπ 2π 2a 2 × 4.5 × 10−2 = → λc = = a kc m m

fc =

3 × 108 3 × 108 × m = Hz = 3.3333m λc 9 × 10−2

m

and GHz

136

ELECTROMAGNETIC FIELDS AND WAVES

TABLE 4.1

Signal Propagation in Parallel-Plate Waveguides TEm0 Modes

TMm0 Modes

mπ a

mπ a

kc

γm0 Hz (x, z)

−

Hx (x, z)

kc H0 cos(kc x)e−γm0 z j ωµ 0

Ez (x, z) −

kc2 − k02

γm0 H0 sin(kc x)e−γm0 z j ωµ

kc2 − k02 0

j E0

kc sin(kc a)e−γm0 z ωε 0

Hy (x, z)

0

E0 cos(kc x)e−γm0 z

Ex (x, z)

0

γm0 E0 cos(kc x)e−γm0 z j ωε

Ey (x, z)

H0 sin(kc x)e−γm0 z

0

Hence, the cutoff frequencies for first four TM modes are found as follows: TM10 → fc = 3.3333 GHz TM20 → fc = 6.6667 GHz TM30 → fc = 10 GHz TM40 → fc = 13.3333 GHz Since the cutoff frequency for the TM40 mode is higher than the signal frequency of 12 GHz, only TM10 , TM20 , and TM30 modes will exist for this signal. The corresponding cutoff wavelengths are 9, 4.5, and 3 cm, respectively. The signal wavelength inside the guide for each mode can be found as follows: λ0 2.5 TM10 → λg = = cm = 2.6024 cm 2 1 − (λ0 /λc ) 1 − (2.5/9)2 λ0 2.5 TM20 → λg = = cm = 3.0067 cm 2 1 − (λ0 /λc ) 1 − (2.5/4.5)2 and λ0 2.5 TM30 → λg = = cm = 4.5227 cm 2 1 − (λ0 /λc ) 1 − (2.5/3)2

137

METALLIC RECTANGULAR WAVEGUIDE

4.8 METALLIC RECTANGULAR WAVEGUIDE In this section we present another application of the formulation described in Section 4.6 to analyze the electromagnetic signals propagating through a hollow metallic cylindrical waveguide of rectangular cross section, as shown in Figure 4.14. Both the geometry of a rectangular waveguide and the coordinate system are shown. It is assumed that its cross section is a × b, and the electromagnetic signal is propagating along the z-axis. Therefore, the field variation along the z-axis is given as e−j kz z . TM and TE modes propagating through this waveguide can be analyzed using equations (4.6.1) to (4.6.18). For TE modes, (4.6.12) reduces to ∂ 2 Fz (x, y) ∂ 2 Fz (x, y) + + (k 2 − kz2 )Fz (x, y) = 0 ∂x 2 ∂y 2

(4.8.1)

It is a second-order partial differential equation that can be solved using the separation-of-variables technique as follows. Assume that Fz (x, y) is a product of two functions, one dependent on x only and the other on y only. Hence, Fz (x, y) = f1 (x)f2 (y)

(4.8.2)

∂ 2 Fz (x, y) d 2 f1 (x) = f2 (y) 2 ∂x dx 2

(4.8.3)

d 2 f2 (y) ∂ 2 Fz (x, y) = f (x) 1 ∂y 2 dy 2

(4.8.4)

Therefore,

and

Substituting (4.8.2) to (4.8.4) into (4.8.1) and rearranging results in 1 d 2 f2 (y) 1 d 2 f1 (x) + = −(k 2 − kz2 ) f1 (x) dx 2 f2 (y) dy 2

(4.8.5)

y

b z

a x

Figure 4.14 Metallic rectangular waveguide.

138

ELECTROMAGNETIC FIELDS AND WAVES

In (4.8.5), the first term is only x dependent, the second is dependent on y only, and the sum of these two is always a constant, as its right-hand side indicates. Therefore, each term on the left-hand side must be independently constant as follows: 1 d 2 f1 (x) = −kx2 f1 (x) dx 2

(4.8.6)

1 d 2 f2 (y) = −ky2 f2 (y) dy 2

(4.8.7)

and

where kx and ky are constants. Equations (4.8.6) and (4.8.7) are in the form of the harmonic equation considered earlier, and its solutions are the harmonic functions. Complete solutions to each of these can be written as f1 (x) = c1 cos kx x + c2 sin kx x

(4.8.8)

f2 (y) = c3 cos ky y + c4 sin ky y

(4.8.9)

and

Therefore, the solution to (4.8.1) is found to be Fz (x, y) = (c1 cos kx x + c2 sin kx x)(c3 cos ky y + c4 sin ky y)

(4.8.10)

Further, equations (4.8.1), (4.8.6), and (4.8.7) give the separation equation, kx2 + ky2 + kz2 = k 2

(4.8.11)

Since tangential electric fields must be zero on the conducting surfaces, the boundary conditions for (4.8.1) are found to be ∂Fz y=0 Ex | = 0 → =0 (4.8.12) y=b ∂y y=0 y=b

and

∂Fz Ey x=0 = 0 → =0 x=a ∂x x=0

(4.8.13)

x=a

To satisfy these boundary conditions, (4.8.10) in conjunction with (4.8.12) and (4.8.14) gives c4 = 0 nπ ky = b

(4.8.14) (4.8.15)

METALLIC RECTANGULAR WAVEGUIDE

c2 = 0

139

(4.8.16)

and kx =

mπ a

(4.8.17)

where m and n are integers, including zero, except that both cannot be zero at the same time (which is a trivial case). Therefore, mπ nπ x cos y (4.8.18) Fz (x, y) = c1 c3 cos a b Further, kz may be found from (4.8.11) as kz =

k2 −

mπ 2 a

−

nπ 2 b

(4.8.19)

Consider a case when the waveguide is air-filled and therefore k = k0 is a pure real number. The following conditions are possible in this situation. (1) k02

mπ 2 a

+

nπ 2 b

(4.8.21)

In this case, the quantity under the square root in (4.8.19) is positive. Therefore, kz is real and the signal propagates through the waveguide without attenuation (ideally). This may be used to transmit the signal. (3) k02 =

mπ 2 a

+

nπ 2 b

= kc2

(4.8.22)

In this case, the quantity under the square root in (4.8.19) goes to zero. As a result, kz is zero (i.e., the cutoff condition). This wave number is known as the cutoff wave number kc , and the corresponding wavelength of the signal is called the cutoff wavelength λc . The corresponding frequency fc is called the cutoff frequency of the waveguide. Thus, the waveguide works as a high-pass filter. A signal propagates only when its frequency is greater than the cutoff frequency of the waveguide.

140

ELECTROMAGNETIC FIELDS AND WAVES

The wavelength and phase velocity of a signal propagating on the waveguide are found as follows:

and

2π 2π λ0 λg = = = 2 k0 1 − (kc /k0 ) 1 − (λ0 /λc )2 k02 − kc2 λ0 = 1 − (fc /f )2 vp =

ω f λ0 3 × 108 = = kz 1 − (λ0 /λc )2 1 − (λ0 /λc )2

m/s

(4.8.23)

(4.8.24)

where f is the signal frequency, k0 its wave number in free space, and λ0 the signal wavelength in free space. Fields inside the waveguide are found after substituting (4.8.18) into (4.6.13) to (4.6.18) as follows: Ex (x, y, z) = j H0 ωµky cos kx x sin ky y e−j kz z Ey (x, y, z) = −j H0 ωµkx sin kx x cos ky y e

−j kz z

Ez (x, y, z) = 0

and

(4.8.25) (4.8.26) (4.8.27)

mπ nπ Hx (x, y, z) = j H0 kz kx sin x cos y e−j kz z a b

(4.8.28)

Hy (x, y, z) = j H0 kz ky cos kx x sin ky ye−j kz z

(4.8.29)

Hz (x, y, z) = H0 kc2 cos kx x cos ky ye−j kz z

(4.8.30)

where H0 = −j c1 c3

1 ωµε

(4.8.31)

It is to be noted that if m and n are both zero, the fields do not exist. Therefore, for a nontrivial case, m or n has to be a nonzero integer. For a > b, the lowestorder mode that propagates is TE10 (i.e., for m = 1 and n = 0). It is known as the dominant mode for rectangular waveguides. From the expressions given in Table 4.2, it may be found easily that the lowest-order TM mode that can propagate through these waveguides is TM11 (i.e., for m = 1 and n = 1). For the TE10 mode, only the Hz , Hx , and Ey field components will be nonzero. Further, λc = 2a and λ0 λg(TE10 ) = (4.8.32) 1 − (λ0 /2a)2 Further,

π π sin x e−j kz z a a π j [(πx/a)+kz z] = −H0 ωµ0 − e−j [(πx/a)+kz z] e 2a

Ey (x, y, z) = −j H0 ωµ0

(4.8.33)

141

METALLIC RECTANGULAR WAVEGUIDE

TABLE 4.2 Section

Signal Propagation in Metallic Waveguides of Rectangular Cross

kc

TEmn Modes mπ 2 nπ 2 + a b

TMmn Modes mπ 2 nπ 2 + a b

k02 − kc2

k02 − kc2

λ0

λ0 1 − (kc /k0 )2

kz = −j γ λg

H0 kc2 cos

Hz (x, y)

Hy (x, y)

nπ mπ x cos y a b

0 nπ mπ x sin y a b mπ nπ nπ j E0 ωε sin x cos y b a b mπ nπ mπ −j E0 ωε cos x sin y a a b E0 kc2 sin

0

Ez (x, y) Hx (x, y)

1 − (kc /k0 )2

mπ nπ mπ sin x cos y a a b mπ nπ nπ cos x sin y j H0 kz b a b

j H0 kz

mπ nπ nπ cos x sin y b a b

−j

nπ mπ mπkz E0 cos x sin y a a b

mπ nπ mπ sin x cos y a a b

−j

nπ mπ nπkz E0 sin x cos y b a b

Ex (x, y)

j H0 ωµ0

Ey (x, y)

−j H0 ωµ0

On comparing this with (4.2.16), it may be found that this represents two plane electromagnetic waves propagating at angles ±θ after reflection from the sidewalls of the waveguide, as shown in Figure 4.15. This angle is found to be θ = sin−1

λ0 2a

(4.8.34)

Therefore, θ → 90◦ as λ0 → 2a (i.e., λc ), and the wave ceases to propagate.

x −θ a θ

z

Figure 4.15 Two uniform plane waves bouncing from the conducting sidewalls of a rectangular waveguide.

142

ELECTROMAGNETIC FIELDS AND WAVES

Example 4.12 The inside cross section of an air-filled rectangular metallic waveguide is 1.58 cm × 0.79 cm. (a) Determine its cutoff frequencies for TE10 , TE20 , TE01 , and TE11 modes. (b) Find the mode(s) that will propagate through this waveguide if the signal frequency is anywhere between 12 and 18 GHz. (c) If this waveguide is being used to send a 15.8-GHz signal in TE10 mode, find the phase velocity. SOLUTION (a) Since kc =

mπ 2 a

+

nπ 2 b

→ fc = 3 × 10

8

m 2 2a

+

n 2 2b

cutoff frequencies for various modes are found to be

2 2 0 1 + Hz = 9.4937 GHz TE10 → fc = 3 × 108 2 × 0.0158 2 × 0.0079

2 2 0 2 8 TE20 → fc = 3 × 10 + Hz = 18.9873 GHz 2 × 0.0158 2 × 0.0079

2 2 1 0 8 TE01 → fc = 3 × 10 + Hz = 18.9873 GHz 2 × 0.0158 2 × 0.0079 and

TE11 → fc = 3 × 108

1 2 × 0.0158

2 +

1 2 × 0.0079

2 Hz = 21.2285 GHz

(b) From part (a), the cutoff frequency for the TE10 mode is 9.4937 GHz. Since the next-higher modes have cutoff at 18.9873 GHz, only the TE10 mode will exist for the signal frequency band 12 to 18 GHz. 3 × 108 3 × 108 3 × 108 (c) vp = = = 1 − (λ0 /λc )2 1 − (fc /f )2 1 − (9.4937/15.8)2 = 3.7531 × 108 m/s 4.9 METALLIC CIRCULAR WAVEGUIDE Electromagnetic signal propagating through a hollow conducting cylindrical waveguide of circular cross section can be analyzed using (4.6.19) to (4.6.36). Figure 4.16 shows the geometry and coordinate systems for such analysis.

143

METALLIC CIRCULAR WAVEGUIDE x

φ a ρ z

y

Figure 4.16 Metallic circular waveguide geometry.

Assume that the waveguide radius is a and the field variation in the z-direction is e−j kz z (because the signal is propagating along z). For a TM mode, (4.6.21) gives ρ

∂ ∂Az (ρ, φ) ∂ 2 Az (ρ, φ) + (k 2 − kz2 )ρ2 Az (ρ, φ) = 0 ρ + ∂ρ ∂ρ ∂φ2

(4.9.1)

Following the technique used in Section 4.8, assume that Az (ρ, φ) is a product of two functions (one dependent on ρ only and the other on φ only) and solve (4.9.1) using the separation of variables. Hence, Az (ρ, φ) = f1 (ρ)f2 (φ)

(4.9.2)

Therefore, (4.9.1) may be written as 1 d 2 f2 (φ) ρ d df1 (ρ) = n2 ρ + (k 2 − kz2 )ρ2 = − f1 (ρ) dρ dρ f2 (φ) dφ2

(4.9.3)

where n is a constant that turns out to be an integer so that f2 (φ) is a single-valued harmonic function. After defining k 2 − kz2 = kρ2 for simplicity, (4.9.3) gives d df1 (ρ) ρ ρ + (kρ2 ρ2 − n2 )f1 (ρ) = 0 dρ dρ

(4.9.4)

d 2 f2 (φ) + n2 f2 (φ) = 0 dφ2

(4.9.5)

and

Equation (4.9.4) is known as Bessel’s equation of order n. Its solutions are the Bessel’s function of the first kind, the Neumann function (also known as the

144

ELECTROMAGNETIC FIELDS AND WAVES

Bessel’s function of the second kind), and the Hankel functions of the first and second kinds. Any two of these functions are linearly independent solutions to (4.9.4). Further, the first two of these represent standing waves, whereas the Hankel functions are convenient for traveling waves. For the present case, only Bessel’s functions of the first kind are possible solutions because Neumann functions go to infinity at the origin (i.e., ρ = 0). Equation (4.9.5) is the harmonic equation considered earlier. Therefore, Az (ρ, φ, z) = cJn (kρ ρ)

sin nφ cos nφ

(4.9.6)

Since tangential electric fields must be zero on the conducting wall at ρ = a, Jn (kρ a) = 0 → kρ =

pnm a

(4.9.7)

where pnm is the mth zero of Jn (kρ a).

TABLE 4.3

Signal Propagation in Circular-Cross-Section Waveguides

kc

TEnm Modes

TMnm Modes

pnm a

pnm a

k02 − kc2

kz = −j γmn

Hz (ρ, φ)

H0 Jn (kc ρ)

cos nφ

k02 − kc2 0

sin nφ

Ez (ρ, φ)

Hρ (ρ, φ)

Hφ (ρ, φ)

Eρ (ρ, φ)

Eφ (ρ, φ)

E0 Jn (kc ρ)

0

−j

−j

−j

kz H0 Jn (kc ρ) kc

nkz H0 Jn (kc ρ) ρkc2

sin nφ

nωµ H0 Jn (kc ρ) ρkc2

ωµ j H0 Jn (kc ρ) kc

cos nφ

− sin nφ

cos nφ − sin nφ

cos nφ cos nφ sin nφ

j

nωε E0 Jn (kc ρ) ρkc2

−j

−j

cos nφ sin nφ − sin nφ

ωε E0 Jn (kc ρ) kc kz E0 Jn (kc ρ) kc

nkz −j 2 E0 Jn (kc ρ) ρkc

cos nφ cos nφ sin nφ

cos nφ sin nφ

− sin nφ cos nφ

145

PROBLEMS

Similarly, (4.6.30) is solved to analyze TE modes in the circular waveguide. In this case, enforcement of the boundary condition (Eφ = 0 at ρ = a) gives ∂Jn (kρ ρ) = Jn (kρ a) = 0 (4.9.8) ∂(kρ ρ) ρ=a and kρ =

pnm a

(4.9.9)

is the mth root of (4.9.8). where pnm Characteristics of the TM and TE modes in a circular waveguide are listed in Table 4.3. Detailed derivations are deferred to Problems 4.34 and 4.35.

SUGGESTED READING Arfken, G., Mathematical Methods for Physicists. San Diego, CA: Academic Press, 1985. Balanis, C. A., Advanced Engineering Electromagnetics. New York: Wiley, 1989. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Harrington, R. F., Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill, 1961. Inan, U. S., and A. S. Inan, Electromagnetic Waves. Upper Saddle River, NJ: Prentice Hall, 2000.

PROBLEMS 4.1. The magnetic flux density on the y = 0 plane is given by B = 5 cos 3x cos 105 t yˆ

T

A square loop of conducting wire is placed on this plane with its vertices at (x, 0, 2), (x, 0, 3), (x + 1, 0, 3), and (x + 1, 0, 2). (a) Find the emf induced around the loop assuming that the loop is stationary. (b) If the loop is moving with a velocity of v = xˆ ž 104 m/s, determine the new induced emf. 4.2. If B = zˆ ž 5 cos(105 t − 0.5πx − πy)T, find the emf induced around a closed loop formed by connecting successively the points (0, 0, 0), (2, 0, 0), (2, 1, 0), (0, 1, 0), and (0, 0, 0). 4.3. For the following charge distribution, find the displacement flux that emanates from a cubical surface bounded by x = ±2, y = ±2, and z = ±2. ρ(x, y, z) = 5(3 − x 2 − y 2 − z2 )

C/m3

146

ELECTROMAGNETIC FIELDS AND WAVES

4.4. Find the charge densities that will produce the following electric flux densities: (a) D = xy xˆ + yzyˆ + zx zˆ C/m2 and (b) D = ρ sin φφˆ C/m2 . 4.5. In cylindrical coordinates, charge is distributed with uniform density 5 C/m3 within the region 1 < ρ < 2. Find the electric flux density in the region (a) 1 < ρ < 2 and (b) ρ > 2. 4.6. If the electric field intensity in free space is E = 5 cos(109 t − x + ky)ˆz V/m, find the value(s) of k. 4.7. The magnetic field intensity of an electromagnetic wave is given as H(x, t) = zˆ ž 0.6 cos βx cos 108 t

A/m

If the region is nonmagnetic with σ = 0 and ε = 6.25ε0 , find the corresponding electric field intensity and β. 4.8. A uniform plane electromagnetic wave is propagating in free space. Its instantaneous magnetic field intensity is given by π µA/m H = zˆ ž 5 cos 108 t − βy + 4 (a) Determine β and the signal wavelength. (b) Find the corresponding instantaneous electric field intensity. 4.9. A uniform plane wave of 50 MHz propagates in a lossless nonmagnetic medium in the +x direction. A probe located at x = 0 measures the phase angle of the wave to be 95◦ . An identical probe located at x = 2.5 m measures the phase to be −12◦ . What is the relative permittivity of the medium? 4.10. The magnetic field intensity of a uniform plane wave of 50 MHz is given as H(z) = (5xˆ + j 10y)e ˆ −j 2z A/m If it is in a lossless and nonmagnetic medium, find vp , η, and the instantaneous electric field intensity. 4.11. The electric field intensity of a uniform plane wave propagating in a nonmagnetic medium of zero conductivity is given as follows: E(y, t) = xˆ ž 12π cos(9π × 107 t + 0.3πy)

V/m

Find (a) the frequency, (b) the wavelength, (c) the direction of wave propagation, (d) the dielectric constant of the medium, and (e) the associated magnetic field intensity. 4.12. A uniform plane wave of 2 GHz is propagating in the +x direction in a nonmagnetic medium that has a dielectric constant of 2.25 and a loss tangent of 0.1. Its electric field is in the y-direction.

147

PROBLEMS

(a) Find the distance over which the amplitude of the propagating wave will reduce by 50 percent. (b) Find the intrinsic impedance, the wavelength, and the phase velocity of the wave in the medium. 4.13. A uniform plane wave propagates in the +z (downward) direction into the ocean (εr = 80, µr = 1, and σ = 4 S/m). Its magnetic field at the ocean surface (z = 0) is give as H(0, t) = yˆ ž 2 cos 109 t A/m (a) Determine the skin depth and intrinsic impedance of the ocean water. (b) Find the corresponding electric field intensity in the ocean.

4.14. A uniform plane wave of frequency 100 kHz is propagating in a material medium. It is given that (1) the fields attenuate by the factor e−1 over a distance of 1205 m, (2) the fields undergo a change in phase by 2π radians in a distance of 1321 m, and (3) the ratio of the amplitudes of its electric and magnetic field intensities at a point in the medium is 163.54 . Find the propagation constant γ and the intrinsic impedance η. 4.15. The electric field intensity of a uniform plane wave propagating through a lossless nonmagnetic medium is given by E(z, t) = xˆ ž 0.2 cos(109 t + πz)

V/m

Find (a) the direction of propagation, (b) the velocity of the signal, (c) the wavelength, and (d) the associated magnetic field intensity. 4.16. A material has an intrinsic impedance of 120 24◦ at 320 MHz. If its relative permeability is 4, find its (a) loss tangent, (b) dielectric constant, (c) conductivity, and (d) complex permittivity. 4.17. The electric field intensity of a uniform plane wave traveling through a material medium (εr = 4, σ = 0.1 S/m, µr = 1) is given by yˆ ž 10eγz V/m. If the signal frequency is 2.45 GHz, find the associated magnetic field intensity and propagation constant γ. 4.18. For a uniform plane wave propagating in the +z direction in a nonmagnetic material medium with σ = 10 S/m and εr = 9, the magnetic field intensity in the z = 0 plane is given by H(z = 0, t) = yˆ ž 0.1 cos(2π × 108 t)

A/m

Find the associated electric field intensity in the medium. 4.19. The magnetic field intensity of a uniform plane wave propagating through a certain nonmagnetic medium is given by H(x, t) = zˆ ž 5 cos(109 t − 6x)

A/m

148

ELECTROMAGNETIC FIELDS AND WAVES

Find (a) the direction of wave travel, (b) the velocity of wave, (c) the wavelength, and (d) the associated electric field intensity. 4.20. A uniform plane wave propagating through a lossless nonmagnetic medium has a power density of 8.05 W/m2 and an rms electric field intensity at 40.2 V/m. Find (a) the intrinsic impedance of the medium, (b) the rms magnetic field intensity, and (c) the velocity of the wave. 4.21. The electric field intensity of a signal transmitted through a coaxial line is given as follows: E = ρˆ

E0 −j βz e ρ

a≤ρ≤b

V/m

where a and b are inner and outer conductor radii, respectively, and β = ω(µε)0.5 . Find (a) the associated magnetic field intensity, (b) the surface current density Js , (c) the surface charge density ρs , and (d) the displacement current per unit length. 4.22. The region x > 0 is a perfect dielectric with εr = 6.25, while the region x < 0 is a perfect dielectric of εr = 2.25. At the interface, subscript 1 denotes field components on the +x side of the boundary and subscript 2 on the −x side. For D1 = 4xˆ + 2yˆ C/m2 , find D2 , E1 , and E2 . 4.23. A sphere of 1 m radius is made of perfect dielectric material (medium 1). It is surrounded by free space (medium 2). The electric field intensities in the two media are given as follows: ˆ E1 = E01 (cos θ rˆ − sin θ θ) and

E2 = E02

8 1+ 3 r

r ≤ 1m

4 cos θ rˆ − 1 − 3 r

sin θ θˆ

r ≥ 1m

Find the permittivity of the spherical medium. 4.24. A 150-MHz plane wave is normally incident from air (region 1) onto a semi-infinite nonmagnetic dielectric slab (region 2). If the voltage standing wave ratio in front of the slab is 2.5 and Emin is at the boundary, find (a) η2 , (b) εr2 , (c) the reflection coefficient, and (d) the distance d from the boundary to the nearest Emax of the standing wave pattern. 4.25. A uniform plane wave traveling through a nonmagnetic dielectric medium with εr = 4 is incident normally upon its interface with free space. If its electric field intensity is given by Ein = yˆ ž 3e−j 200πz

mV/m

149

PROBLEMS

Find (a) the corresponding magnetic field intensity, (b) the reflected electric field intensity, (c) the reflected magnetic field intensity, (d) the transmitted electric field intensity, and (e) the transmitted magnetic field intensity. 4.26. A uniform plane wave of 3 GHz propagating through free space is incident normally on a lossless dielectric medium with εr = 4 and µr = 1. The incident electric field at the interface has a value of 2 mV/m right before it strikes the interface. In free space, find (a) the reflection coefficient, (b) the VSWR, (c) the positions (in meters) of the maxima and minima of the electric field, and (d) the maximum and minimum values of the electric field. 4.27. The electric field intensity of a uniform plane wave propagating in air is Ei (z) = xˆ ž 5e−j 6z V/m. The wave is incident normally on an interface at z = 0 with a medium that has a dielectric constant at 6.25 and a loss tangent at 0.3. Find (a) the phasor expressions for reflected and transmitted electric and magnetic fields, and (b) the time-averaged power flow per unit area in the lossy medium. 4.28. Region 1 (x < 0) is air, whereas region 2 (x > 0) is a nonmagnetic medium characterized by σ = 10−4 S/m and εr = 4. A uniform plane wave with its electric field intensity as given below is incident on the interface at x = 0 from region 1.

2π x Ein = 5 cos 2π × 10 t − 3 8

V/m

Obtain the reflected and transmitted wave electric fields. 4.29. Verify the characteristics of the TE mode fields given in Table 4.1 for an electromagnetic signal that propagates in a metallic parallel-plate waveguide. 4.30. Separation between the plates of a parallel-plate waveguide is 0.625 cm, filled with a nonmagnetic dielectric material of εr = 4. Find the cutoff frequencies for the TM00 , TM10 , TM20 , TE10 , and TE20 modes. Can a signal propagate in the TE00 mode? Determine the phase velocities for all possible TE and TM modes that exist at 30 GHz. 4.31. Verify the characteristics of TM mode fields given in Table 4.2 for an electromagnetic signal that propagates in a metallic waveguide of rectangular cross section. 4.32. The wavelength of a signal propagating through an air-filled WR-340 rectangular waveguide is found to be 20 cm. Determine its frequency in gigahertz. 4.33. The inside cross section of an air-filled rectangular waveguide is 2.286 cm × 1.016 cm.

150

ELECTROMAGNETIC FIELDS AND WAVES

(a) Determine its cutoff frequencies for the TE10 , TE20 , TE01 , and TE11 modes. (b) Which mode(s) will propagate through this waveguide if the signal frequency is anywhere between 8 and 12 GHz? (c) Determine the phase velocity of a 9.375 GHz signal propagating in TE10 mode. 4.34. Verify the characteristics of TE mode fields given in Table 4.3 for an electromagnetic signal that propagates in a hollow metallic cylindrical waveguide of circular cross section. 4.35. Verify the characteristics of the TM mode fields given in Table 4.3 for an electromagnetic signal that propagates in a hollow metallic cylindrical waveguide of circular cross section.

5 RESONANT CIRCUITS

A communication circuit designer frequently requires means to select (or reject) a band of frequencies from a wide signal spectrum. Resonant circuits provide such filtering. There are well-developed, sophisticated methodologies to meet virtually any specification. However, a simple circuit suffices in many cases. Further, resonant circuits are an integral part of the frequency-selective amplifier as well as of the oscillator designs. These networks are also used for impedance transformation and matching. In this chapter we describe the analysis and design of these simple frequencyselective circuits and present the characteristic behavior of series and parallel resonant circuits. Related parameters, such as quality factor, bandwidth, and input impedance, are introduced that will be used in several subsequent chapters. Transmission lines with an open or short circuit at their ends are considered next, and their relationships with the resonant circuits are established. Transformer-coupled parallel resonant circuits are discussed briefly because of their significance in the radio-frequency range. In the final section we summarize the design procedure for rectangular and circular cylindrical cavities, and the dielectric resonator.

5.1 SERIES RESONANT CIRCUITS Consider the series R –L–C circuit shown in Figure 5.1. Since the inductive reactance is directly proportional to signal frequency, it tries to block the highfrequency contents of the signal. On the other hand, capacitive reactance is Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

151

152

RESONANT CIRCUITS L

C

vin(t)

R

vo(t)

Figure 5.1 Series R –L–C circuit with input–output terminals.

inversely proportional to the frequency. Therefore, it tries to stop its lower frequencies. Note that the voltage across an ideal inductor leads the current by 90◦ (i.e., the phase angle of an inductive reactance is 90◦ ). In the case of a capacitor, voltage across its terminals lags behind the current by 90◦ (i.e., the phase angle of a capacitive reactance is −90◦ ). This means it is possible that the inductive reactance will be canceled out by the capacitive reactance at some intermediate frequency. This frequency is called the resonant frequency of the circuit. If the input signal frequency is equal to the resonant frequency, maximum current will flow through the resistor and it will be in phase with the input voltage. In this case, the output voltage Vo will be equal to the input voltage Vin . It can be analyzed as follows: From Kirchhoff’s voltage law, t L dvo (t) 1 vo (t) dt + vo (t) = vin (t) (5.1.1) + R dt RC −∞ Taking the Laplace transform of this equation with initial conditions as zero (i.e., no energy storage initially), we get sL 1 (5.1.2) + + 1 Vo (s) = Vin (s) R sRC where s is the complex frequency (Laplace variable). The transfer function of this circuit, T (s), is given by T (s) =

Vo (s) 1 sR = = 2 Vin (s) sL/R + 1/sRC + 1 s L + sR + 1/C

(5.1.3)

Therefore, the transfer function of this circuit has a zero at the origin of the complex s-plane and also has two poles. The location of these poles can be determined by solving the quadratic equation s 2 L + sR +

1 =0 C

Two possible solutions to this equation are as follows: 2 R 1 R − ± s1,2 = − 2L 2L LC

(5.1.4)

(5.1.5)

SERIES RESONANT CIRCUITS

153

The circuit response will be influenced by the location of these poles. Therefore, these networks can be characterized as follows: √ √ 1. If R/2L > 1/ LC (i.e., R > 2 L/C), both of these poles will be real and distinct, and the circuit is overdamped. √ √ 2. If R/2L = 1/ LC (i.e., R = 2 L/C), the transfer function will have √ double poles at s = −R/2L = −1/ LC. The circuit is critically damped. √ √ 3. If R/2L < 1/ LC (i.e., R < 2 L/C), the two poles of T (s) will be complex conjugate of each other. The circuit is underdamped. Alternatively, the transfer function may be rearranged as follows: T (s) =

sCRω20 sCR = s 2 LC + sRC + 1 s 2 + 2ζω0 s + ω20

(5.1.6)

where R ζ= 2 ω0 = √

C L

1 LC

(5.1.7) (5.1.8)

ζ is called the damping ratio and ω0 is the undamped natural frequency. Poles of T (s) are determined by solving the equation s 2 + 2ζω0 s + ω20 = 0

(5.1.9)

For ζ < 1, s1,2 = −ζω0 ± j ω0 1 − ζ2 . As shown in Figure 5.2, the two poles are complex conjugates of each other. The output transient response will be oscillatory with a ringing frequency of ω0 (1 − ζ2 ) and an exponentially decaying amplitude. This circuit is underdamped. For ζ = 0, the two poles move on the imaginary axis. Transient response will be oscillatory. It is a critically damped case. For ζ = 1, the poles are on the negative real axis. Transient response decays exponentially. In this case, the circuit is overdamped. Consider the unit step function shown in Figure 5.3. It is like a direct voltage source of 1 V that is turned on at time t = 0. If it represents input voltage vin (t), the corresponding output vo (t) can be determined via Laplace transform technique. The Laplace transform of a unit step at the origin is equal to 1/s. Hence, the output voltage, vo (t), is found as follows: vo (t) = L−1 Vo (s) = L−1

sCRω20 CRω20 1 −1 = L s 2 + 2ζω0 s + ω20 s (s + ζω0 )2 + (1 − ζ2 )ω20

154

RESONANT CIRCUITS jw

w0(1 − ζ2)1/2

w0

−σ −w0(1 − ζ2)1/2

−ζw0

Figure 5.2 Pole–zero plot of the transfer function.

vin(t)

1V

Time

t→∞

t=0

Figure 5.3

Unit-step input voltage.

where L−1 represents the inverse Laplace transform operator. Therefore, 2ζ vo (t) = e−ζω0 t sin(ω0 t 1 − ζ2 )u(t) 1 − ζ2 This response is illustrated in Figure 5.4 for three different damping factors. As can be seen, initial ringing lasts longer for a lower damping factor. A sinusoidal steady-state response of the circuit can be determined easily after replacing s by j ω, as follows: Vo (j ω) =

Vin (j ω) Vin (j ω) = j ωL/R + 1/j ωRC + 1 (1 + j/RC)(LCω − 1/ω)

Vo (j ω) =

Vin (j ω) Vin (j ω) = 2 (1 + j/ω RC)(ω/ω (1 + j/RC)(ω/ω0 − 1/ω) 0 0 − ω0 /ω)

or

155

SERIES RESONANT CIRCUITS 0.8

ζ = 0.95

0.6

Output voltage

0.4 ζ = 0.5

0.2

ζ = 0.1

0.0

−0.2

10

0

20 w0t

30

40

50

Figure 5.4 Response of a series R –L–C circuit to a unit step input for three different damping factors.

The quality factor, Q, of the resonant circuit is a measure of its frequency selectivity. It is defined as Q = ω0

average stored energy power loss

(5.1.10)

Hence, 1

Q = ω0 12 2

LI 2 I 2R

=

ω0 L R

Since ω0 L = 1/ω0 C, √ ω0 L LC 1 1 L 1 Q= = = = = R ω0 RC RC R C 2ζ Therefore, Vo (j ω) =

Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω)

(5.1.11)

(5.1.12)

Alternatively, 1 Vo (j ω) = A(j ω) = Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω)

(5.1.13)

The magnitude and phase angle of (5.1.13) are illustrated in Figures 5.5 and 5.6, respectively. Figure 5.5 shows that the output voltage is equal to the input for

156

RESONANT CIRCUITS

1.0

Q=1

0.8

Magnitude

0.6

Q = 10

0.4 Q = 100 0.2 ω0 0.0 0.0

20.0

40.0

60.0

80.0

100.0

ω

Figure 5.5

Magnitude of A(j ω) as a function of ω.

2.0 Q = 100

1.5

Phase angle (rad)

1.0 0.5 Q = 10

0.0

Q=1

−0.5 −1.0 −1.5 −2.0 0.0

ω0 20.0

40.0

60.0

80.0

100.0

ω

Figure 5.6 Phase angle of A(j ω) as a function of ω.

a signal frequency equal to the resonant frequency of the circuit. Further, phase angles of the two signals in Figure 5.6 are the same at this frequency, irrespective of the quality factor of the circuit. As signal frequency moves away from this point on either side, the output voltage decreases. The rate of decrease depends on the quality factor of the circuit. For higher Q, the magnitude is sharper,

157

SERIES RESONANT CIRCUITS

indicating a higher selectivity of the circuit. If signal frequency is below the resonant frequency, the output voltage leads the input. For a signal frequency far below the resonance, the output leads the input by almost 90◦ . On the other hand, it lags behind the input for higher frequencies. It converges to −90◦ as the signal frequency moves far beyond the resonant frequency. Thus, the phase angle changes between π/2 and −π/2, following a sharper change around the resonance for high-Q circuits. Note that the voltage across the series-connected inductor and capacitor combined has inverse characteristics to those of the voltage across the resistor. Mathematically, VLC (j ω) = Vin (j ω) − Vo (j ω) where VLC (j ω) is the voltage across the inductor and capacitor combined. In this case, sinusoidal steady-state response can be obtained as follows: VLC (j ω) Vo (j ω) 1 =1− =1− Vin (j ω) Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω) =

j Q(ω/ω0 − ω0 /ω) 1 + j Q(ω/ω0 − ω0 /ω)

Hence, this configuration of the circuit represents a band-rejection filter. Half-power frequencies ω1 and ω2 of a bandpass circuit can be determined from (5.1.13) as 1 1 = ⇒ 2 = 1 + Q2 2 2 1 + Q (ω/ω0 − ω0 /ω)2 Therefore,

ω0 ω − Q ω0 ω

= ±1

Assuming that ω1 < ω0 < ω2 ,

ω1 ω0 Q − ω0 ω1

= −1

and

ω0 ω2 − Q ω0 ω2 Therefore,

=1

ω0 ω1 ω0 ω2 − =− − ω0 ω2 ω0 ω1

ω0 ω − ω0 ω

2

158

RESONANT CIRCUITS

or ω2 ω2 ω2 ω2 ω2 − 0 = −ω1 + 0 ⇒ (ω2 + ω1 ) = 0 + 0 = ω20 ω2 ω1 ω1 ω1

1 1 + ω1 ω2

or ω20 = ω1 ω2

(5.1.14)

and ω2 ω0 1 ω0 ω1 − = − ⇒ ω1 − 0 = − ω0 ω1 Q ω1 Q or ω1 − ω2 = −

ω0 ω0 ⇒Q= Q ω2 − ω1

(5.1.15)

Example 5.1 Determine the element values of a resonant circuit that passes all the sinusoidal signals from 9 to 11 MHz. This circuit is to be connected between a voltage source with negligible internal impedance and a communication system with its input impedance at 50 . Plot its characteristics in a frequency band of 1 to 20 MHz. SOLUTION From (5.1.14), ω0 =

√

ω1 ω2 → f0 =

√ f1 f2 = 9 × 11 = 9.949874 MHz

From (5.1.11) and (5.1.15), Q=

ω0 R ω0 L = ⇒L= R ω1 − ω2 ω1 − ω2 50 = 3.978874 × 10−6 H ≈ 4 µH = 2π × 106 × (11 − 9)

From (5.1.8), ω0 = √

1 LC

⇒C=

1 = 6.430503 × 10−11 F ≈ 64.3 pF Lω20

The circuit arrangement is shown in Figure 5.7. Its magnitude and phase characteristics are displayed in Figure 5.8. Input Impedance Impedance across the input terminals of a series R –L–C circuit can be determined as follows: ω20 1 Zin = R + j ωL + (5.1.16) = R + j ωL 1 − 2 j ωC ω

159

SERIES RESONANT CIRCUITS L = 4 µH

C = 64.3 pF

R = 50 Ω

v

Figure 5.7

Filter circuit arrangement for Example 5.1.

1.0

Magnitude of A(jω)

0.8

0.6

0.4

0.2

0.0

0

5

10 ω (a)

15

0

5

10 ω (b)

15

20

2.0

Phase angle in (rad)

1.5 1.0 0.5 0.0 −0.5 −1.0 −1.5 −2.0

20

Figure 5.8 Magnitude (a) and phase (b) plots of A(j ω) for the circuit in Figure 5.7.

160

RESONANT CIRCUITS

At resonance, the inductive reactance cancels out the capacitive reactance. Therefore, the input impedance reduces to the total resistance of the circuit. If signal frequency changes from the resonant frequency by ±δω, the input impedance can be approximated as Zin = R + j ωL

(ω + ω0 )(ω − ω0 ) 2QRδω ≈ R + j 2δωL = R + j (5.1.17) 2 ω ω0

Alternatively, Zin ≈ R + j 2δωL =

ω0 L ω0 + j 2(ω − ω0 )L = j 2 ω − ω0 + L Q j 2Q

1 L = j 2 ω − ω0 1 + j 2Q

(5.1.18)

Therefore, a series resonant circuit can be analyzed with R as zero (i.e., assuming that the circuit is lossless). The losses can be included subsequently by replacing a real resonant frequency, ω0 , by the complex frequency, ω0 [1 + j (1/2Q)]. At resonance, the current through the circuit, Ir , is Ir =

Vin R

(5.1.19)

Therefore, voltages across the inductor, VL , and the capacitor, VC , are VL = j ω0 L

Vin = j QVin R

(5.1.20)

and VC =

1 Vin = −j QVin j ω0 C R

(5.1.21)

Hence, the magnitude of voltage across the inductor is equal to the quality factor times input voltage, while its phase leads by 90◦ . The magnitude of the voltage across the capacitor is the same as that across the inductor. However, it is 180◦ out of phase because it lags behind the input voltage by 90◦ .

5.2 PARALLEL RESONANT CIRCUITS Consider an R –L–C circuit in which the three components are connected in parallel, as shown in Figure 5.9. A subscript p is used to differentiate the circuit elements from those used in series circuit of Section 5.1. A current source, iin (t), is connected across its terminals and io (t) is current through the resistor Rp .

161

PARALLEL RESONANT CIRCUITS io(t) iin(t)

Lp

Figure 5.9

Cp

Rp

vo(t)

Parallel R –L–C circuit.

Voltage across this circuit is vo (t). From Kirchhoff’s current law, iin (t) =

1 Lp

t −∞

Rp io (t) dt + Cp

d(Rp io (t)) + io (t) dt

(5.2.1)

Assuming that no energy was stored in the circuit initially, we take the Laplace transform of (5.2.1). It gives Iin (s) = Hence,

Rp + sRp Cp + 1 Io (s) sLp

Io (s) sLp /Rp = 2 Iin (s) s Lp Cp + s(Lp /Rp ) + 1

(5.2.2)

Note that this equation is similar to (5.1.6). It changes to T (s) if RC replaces Lp /Rp . Therefore, the results of the series resonant circuit can be used for this parallel resonant circuit, provided that ζ=

1 2 ω0 Rp Cp

and 1 ω0 = Lp Cp Hence, 1 1 = ζ= 2 ω0 Rp Cp 2Rp

(5.2.3) Lp Cp

(5.2.4)

The quality factor, Qp , and the impedance, Zp , of the parallel resonant circuit can be determined as follows: Qseries = Zp =

ω0 L Rp 1 = ⇒ Qp = ω0 Rp Cp = R ω0 RC ω0 Lp

(5.2.5)

I0 (j ω)Rp j ωLp Vo (j ω) = = (5.2.6) 2 Iin (iω) Iin (j ω) −ω Lp Cp + j ω(Lp /Rp ) + 1

162

RESONANT CIRCUITS

Input Admittance Admittance across input terminals of the parallel resonant circuit (i.e., the admittance seen by the current source) can be determined as follows: Yin =

ω2 1 1 1 1 = + j ωCp + = + j ωCp 1 − 02 Zp Rp j ωLp Rp ω

(5.2.7)

Hence, input admittance will be equal to 1/Rp at the resonance. It will become zero (i.e., the impedance will be infinite) for a lossless circuit. It can be approximated around the resonance, ω0 ± δω, as Yin ≈

1 1 2δωQ + j 2δωCp = +j Rp Rp ω0 Rp

(5.2.8)

The corresponding impedance is Zp ≈

Rp 1 + j (2Qδω/ω0 )

(5.2.9)

Current through the capacitor, Ic , at the resonance is Ic = j ω0 Cp Rp Iin = j QIin

(5.2.10)

and current through the inductor, IL , is IL =

1 Rp Iin = −j QIin j ω0 Lp

(5.2.11)

Thus, current through the inductor is equal in magnitude but opposite in phase to that through the capacitor. Further, these currents are larger than the input current by a factor of Q. Quality Factor of a Resonant Circuit If resistance R represents losses in the resonant circuit, Q given by the preceding formulas is known as the unloaded Q. If the power loss due to external load coupling is included through an additional resistance RL , the external Qe is defined as follows: ω L 0 RL Qe = R L ω0 Lp

for series resonant circuit (5.2.12) for parallel resonant circuit

163

PARALLEL RESONANT CIRCUITS

The loaded Q, QL , of a resonant circuit includes internal losses as well as the power extracted by the external load. It is defined as follows: ω0 L RL + R for series resonant circuit

QL = (5.2.13)

Rp R L for parallel resonant circuit ω0 Lp where

RL

Rp =

RL Rp RL + Rp

Hence, the following relation holds good for both types of resonant circuit: 1 1 1 = + QL Qe Q

(5.2.14)

See Table 5.1. Example 5.2 Consider the loaded parallel resonant circuit in Figure 5.10. Compute the resonant frequency in radians per second, unloaded Q, and the loaded Q of this circuit. SOLUTION

TABLE 5.1

1 1 ω0 = =√ = 108 rad/s Lp Cp 10−5 × 10−11 Relations for Series and Parallel Resonant Circuits Series

Parallel

1

LC R C 2 L

1 Lp Cp Lp 1 2Rp Cp

Unloaded Q

1 ω0 L = R ω0 RC

Rp = ω0 Rp Cp ω0 Lp

External Q = Qe

1 ω0 L = RL ω0 RL C

RL = ω0 RL Cp ω0 Lp

Loaded Q = QL

QQe Q + Qe

QQe Q + Qe

ω0 Damping factor, ζ

Input impedance, Zin , around resonance

√

R+j

2RQδω ω0

R 1 + j (2Qδω/ω0 )

164

RESONANT CIRCUITS

Load 10 pF

10 µH

100 kΩ

100 kΩ

Figure 5.10

The unloaded

Circuit for Example 5.2.

Q=

105 Rp = 8 = 100 ω0 Lp 10 × 10−5

Qe =

RL 105 = 8 = 100 ω0 Lp 10 × 10−5

The external Q,

The loaded Q,

Rp

RL QQe 50 × 103 QL = = = 8 = 50 ω0 Lp Q + Qe 10 × 10−5

5.3 TRANSFORMER-COUPLED CIRCUITS Transformers are used as a means of coupling as well as of impedance transforming in electronic circuits. Transformers with tuned circuits in one or both of their sides are employed in voltage amplifiers and oscillators operating at radio frequencies. In this section we present an equivalent model and an analytical procedure for the transformer-coupled circuits. Consider a load impedance ZL that is coupled to the voltage source Vs via a transformer as illustrated in Figure 5.11. Source impedance is assumed to be Zs . The transformer has a turns ratio of n : 1 between its primary (the source) and secondary (the load) sides. Using the notations indicated, equations for various voltages and currents can be written in phasor form as follows: V1 = j ωL1 I1 + j ωMI2

(5.3.1)

V2 = j ωMI1 + j ωL2 I2

(5.3.2)

where M is the mutual inductance between the two sides of the transformer. Standard convention with a dot on each side is used. Hence, magnetic fluxes reinforce each other for the case of currents entering this terminal on both sides, and M is positive.

165

TRANSFORMER-COUPLED CIRCUITS I1

I2

+

+

Zs V1 Vs

L1

ZL

V2

L2 n:1

−

Figure 5.11

−

Transformer-coupled circuit.

The following relations hold for an ideal transformer operating at any frequency: V1 = nV2 I2 I1 = − n and

(5.3.3) (5.3.4)

nV2 V2 V1 = n2 = Z1 = = n2 Z2 I1 −I2 /n −I2

(5.3.5)

Several equivalent circuits are available for a transformer. We consider one of these that is most useful in analyzing the communication circuits. This equivalent circuit is illustrated in Figure 5.12. The following equations for phasor voltages and currents may be formulated using the notations indicated in the figure. I2 I2 V1 = j ω(1 − ξ)L1 I1 + j ωξL1 I1 + (5.3.6) = j ωL1 I1 + j ωξL1 n n I1 + Zs

I2/n

I2 +

(1 − ξ) L1 n:1

V1

V2

ξ L1

ZL

Vs −

− Ideal transformer

Figure 5.12 Equivalent model of the transformer-coupled circuit shown in Figure 5.11.

166

RESONANT CIRCUITS

and

1 I2 V2 = j ωξL1 I1 + n n

(5.3.7)

If the circuit shown in Figure 5.12 is equivalent to that shown in Figure 5.11, these two equations represent the same voltages as those of (5.3.1) and (5.3.2). Hence, ξL1 =M (5.3.8) n and

ξL1 = L2 n2

In other words,

n=

and ξ=

(5.3.9)

ξL1 L2

(5.3.10)

√ M ξ M nM =√ ⇒ ξ= √ =κ L1 L1 L2 L1 L2

(5.3.11)

where κ is called the coefficient of coupling. It is close to unity for a tightly coupled transformer, and close to zero for a loose coupling. Example 5.3 A tightly coupled transformer is used in the circuit shown in Figure 5.13. Inductances of its primary and secondary sides are 320 and 20 nH, respectively. Find its equivalent circuit, the resonant frequency, and the Q of this circuit.

I1 +

6 pF Is

V1

−

Figure 5.13

I2 +

L1

V2

30.7 pF

L2

−

Transformer-coupled resonant circuit.

62.5 kΩ

167

TRANSFORMER-COUPLED CIRCUITS

+

Is

+

V1

R1

C1

V2

L1 n:1

−

− Ideal transformer

Figure 5.14 Equivalent model of the transformer-coupled circuit shown in Figure 5.13.

SOLUTION Since the transformer is tightly coupled, κ ≈ 1. Therefore, ξ ≈ 1, 1 − ξ ≈ 0, and its equivalent circuit simplifies as shown in Figure 5.14. From (5.3.10), ξL1 320 n= =4 ≈ L2 20 and from (5.3.5), Y2 1 Z1 = n Z2 ⇒ Y1 = 2 = 2 n n 2

1 + j ωC2 R2

Therefore, R1 = n2 R2 = 16 × 62.5 k = 1 M 1 C1 = 6 + 2 30.7 pF = 6 + 1.91875 ≈ 7.92 pF 4 1 1 =√ = 628.1486 × 106 rad/s ω0 = √ −9 L1 C1 320 × 10 × 7.92 × 10−12 ω0 = 99.97 MHz ≈ 100 MHz f0 = 2π and Q=

106 R1 = = 4974.9375 ω0 L1 628.1486 × 106 × 320 × 10−9

Example 5.4 A transformer-coupled circuit is shown in Figure 5.15. Draw its equivalent circuit using an ideal transformer. (a) If the transformer is tuned only to its secondary side (i.e., R1 and C1 are removed), determine its resonant frequency and impedance seen by the current source.

168

RESONANT CIRCUITS I1

I2

+

R1

C1

Is

V1

+

C2

V2

L1

L2

R2

−

−

Figure 5.15

I0

Double-tuned transformer-coupled circuit.

(b) Determine the current transfer function (I0 /Is ) for the entire circuit (i.e., R1 and C1 are included in the circuit). If the transformer is loosely coupled, and both of the sides have identical quality factors as well as the resonant frequencies, determine the locations of the poles of the current transfer function on the complex ω-plane. SOLUTION Following the preceding analysis and Figure 5.12, the equivalent circuit can be drawn easily, as shown in Figure 5.16. Using notations as indicated in the figure, circuit voltages and currents can be found as follows: I2 nV2 = sξL1 I1 + (5.3.12) n R2 = R2 I0 1 + sR2 C2 1 + sC1 [nV2 + s(1 − ξ)L1 I1 ] Is = I1 + R1

V2 = −I2

I1 +

Is

C1

−

ξL1

(5.3.14)

I2

+

(1 − ξ)L1 R1

V1

I2/n

(5.3.13)

+

I0

n:1 nV2

V2

−

C2

−

Ideal transformer

Figure 5.16 Equivalent representation for the circuit shown in Figure 5.15.

R2

169

TRANSFORMER-COUPLED CIRCUITS

From (5.3.12), (5.3.13), and (5.3.10), we find that sL2 sξL1 s 2 ξL1 R2 C2 + sξL1 2 I1 = 1 + s L C + V R2 I0 = 1+ 2 2 2 2 n R2 R2 n or

sξL1 /nR2 I0 = 2 I1 1 + s L2 C2 + sL2 /R2

(5.3.15)

When R1 and C1 are absent, I1 will be equal to Is , and (5.3.15) will represent the current transfer characteristics. This equation is similar to (5.2.2). Hence, the resonant frequency is found as follows: ω0 = √

1 L2 C2

(5.3.16)

Note that the resonant frequency in this case depends only on L2 and C2 . It is independent of inductance L1 of the primary side. The impedance seen by the current source, with R1 and C1 removed, is determined as follows: Zin (s) =

V1 n2 sL2 = sL1 (1 − ξ) + 2 I1 s L2 C2 + s(L2 /R2 ) + 1

(5.3.17)

Zin (j ω0 ) = j ω0 L1 (1 − ξ) + n2 R2

(5.3.18)

At resonance, and if transformer is tightly coupled, ξ ≈ 1, Zin (j ω0 ) = n2 R2

(5.3.19)

When R1 and C1 are included in the circuit, (5.3.12) to (5.3.14) can be solved to obtain a relationship between I0 and Is . The desired current transfer function can be obtained as I0 sξL1 /nR2 = 2 Is [s L2 C2 + s(L2 /R2 ) + 1]{1 + s(1 − ξ)L1 [(1/R1 ) + sC1 ]} +sξL1 [(1/R1 ) + sC1 ]

(5.3.20)

Note that the denominator of this equation is now a polynomial of the fourth degree. Hence, this current ratio has four poles on the complex ω-plane. If the transformer is loosely coupled, ξ will be negligible. In that case, the denominator of (5.3.20) can be approximated as follows: I0 sξL1 /nR2 ≈ 2 Is [s L2 C2 + s(L2 /R2 ) + 1]{1 + sL1 [(1/R1 ) + sC1 ]} + sξL1 [(1/R1 ) + sC1 ] (5.3.21)

170

RESONANT CIRCUITS

For ω1 = ω2 = ω0 , and Q1 = Q2 = Q0 , L1 C1 = L2 C2 =

1 ω20

and R1 R2 = = ω0 Q0 L1 L2 Hence, (5.3.21) may be written as sξL1 /nR2 I0 ≈ 2 2 Is (s /ω0 + s/ω0 Q + 1)2 + ξ(s 2 /ω20 + s/ω0 Q)

(5.3.22)

The poles of (5.3.22) are determined by solving the equation

2 2 s2 s s s + + 1 + ξ + =0 ω0 Q0 ω0 Q0 ω20 ω20

(5.3.23)

The roots of this equation are found to be 1 2 ω0 ± j ω0 (1 ∓ ξ) − s≈− 2Q0 2Q0

(5.3.24)

5.4 TRANSMISSION LINE RESONANT CIRCUITS Transmission lines with open- or short-circuited ends are frequently used as resonant circuits in the ultrahigh frequency and microwave range. We consider such networks in this section. Since the quality factor is an important parameter of these circuits, we need to include the finite (even though small) loss in the line. There are four basic types of these networks, as illustrated in Figure 5.17. It can be found easily by analyzing the input impedance characteristics around the resonant wavelength, λr , that the circuits of Figure 5.17 (a) and (d) behave like a series R –L–C circuit. On the other hand, the other two transmission lines possess the characteristics of a parallel resonant circuit. A quantitative analysis of these circuits is presented below for n as unity. Short-Circuited Line Consider a transmission line of length l and characteristic impedance Z0 . It has a propagation constant, γ = α + j β. The line is short circuited at one of its ends, as shown in Figure 5.18. The impedance at its other end, Zin , can be determined

171

TRANSMISSION LINE RESONANT CIRCUITS (2n − 1)λr 4

l=

l=

γ = α + jβ Open Z0

Z0

Open γ = α + jβ

(a) l=

n λr 2

(b)

(2n − 1)λr 4

l=

γ = α + jβ

n λr 2

Z0

Short

Short γ = α + jβ

Z0 (c)

(d )

Figure 5.17 Four basic types of transmission line resonant circuits, n = 1, 2, . . . .

l

Zin →

Z0

γ = α + jβ

Figure 5.18 Short-circuited lossy transmission line.

from (3.2.5) as follows: Zin = Z0 tanh γl = Z0 tanh(α + j β)l = Z0

tanh αl + j tan βl 1 + j tanh αl tan βl

(5.4.1)

For αl 1, tanh αl ≈ αl, and assuming that the line supports only the TEM mode, we find that βl = ωl/vp . Hence, it can be simplified around the resonant frequency, ω0 , as follows: βl =

ωl ω0 l δωl δωl = + = βr l + vp vp vp vp

(5.4.2)

where βr is the phase constant at the resonance. If the transmission line is one-half wavelength long at the resonant frequency, βr l = π and

l π = . vp ω0

172

RESONANT CIRCUITS

Therefore, tan βl can be approximated as follows: δω l π δω π δω π δω tan βl = tan π + ≈ = tan π + = tan vp ω0 ω0 ω0 and Zin ≈ Z0

αl + j (π δω/ω0 ) π δω ≈ Z0 αl + j 1 + j (αl)(π δω/ω0 ) ω0

(5.4.3)

It is assumed in (5.4.3) that (αl)(π δω/ω0 ) 1. For a series resonant circuit, Zin ≈ R + j 2δωL. Hence, a half-wavelength-long transmission line with short-circuit termination is similar to a series resonant circuit. The equivalent-circuit parameters are found as follows (assuming that losses in the line are small such that the characteristic impedance is a real number): R ≈ Z0 αl = 12 Z0 αλr

(5.4.4)

π Z0 2 ω0 2 C≈ πω0 Z0 L≈

(5.4.5) (5.4.6)

and Q=

π βr ω0 L ≈ = R 2αl 2α

(5.4.7)

On the other hand, if the transmission line is a quarter-wavelength long at the resonant frequency, βr l = π/2, and l/vp = π/2 ω0 . Therefore,

π δωl tan βl = tan + 2 vp and Zin ≈ Z0

π πδω = tan + 2 2 ω0

πδω = − cot 2 ω0

Z0 /αl −j (2 ω0 /πδω) = 1 − j αl(2 ω0 /πδω) 1 + j (π/αl)(δω/2 ω0 )

≈−

2 ω0 πδω (5.4.8) (5.4.9)

This expression is similar to the one obtained for a parallel resonant circuit in (5.2.9). Therefore, this transmission line is working as a parallel resonant circuit with equivalent parameters as follows: Rp ≈

Z0 4Z0 = αl αλr 4Z0 Lp ≈ πω0 π Cp ≈ 4 ω0 Z0

(5.4.10) (5.4.11) (5.4.12)

TRANSMISSION LINE RESONANT CIRCUITS

173

and Q=

βr π = 4αl 2α

(5.4.13)

Open-Circuited Line The analysis of the open-circuited transmission line shown in Figure 5.19 can be performed following a similar procedure. These results are summarized below. From (3.2.5), Z0 Zin |ZL =∞ = (5.4.14) tanh γl For a transmission line that is a half-wavelength long, the input impedance can be approximated as follows: Zin ≈ Z0

1 + j (αl)(πδω/ω0 ) Z0 ≈ αl + j (πδω/ω0 ) αl + j (πδω/ω0 )

(5.4.15)

This is similar to (5.2.9). Therefore, a half-wavelength-long open-circuited transmission line is similar to a parallel resonant circuit with equivalent parameters as follows: Z0 2Z0 = αl αλr 2Z0 Lp ≈ πω0 π Cp ≈ 2 ω0 Z0

Rp ≈

(5.4.16) (5.4.17) (5.4.18)

and Q=

βr π = 2αl 2α

(5.4.19)

If the line is only a quarter-wavelength long, the input impedance can be approximated as 1 − j (αl)(2 ω0 /πδω) πδω Zin ≈ Z0 (5.4.20) ≈ Z0 αl + j αl − j (2 ω0 /πδω) 2 ω0 l

Zin →

Z0

γ = α + jβ

Figure 5.19 Open-circuited transmission line.

174 TABLE 5.2

RESONANT CIRCUITS

Equivalent-Circuit Parameters of the Resonant Lines for n = 1 Quarter-wavelength line

Resonance R L C Q

Half-wavelength line

Open Circuit

Short Circuit

Open Circuit

Short Circuit

Series

Parallel

Parallel

Series

4Z0 αλr 4Z0 πω0 π 4Z0 ω0 βr 2α

2Z0 αλr 2Z0 πω0 π 2Z0 ω0 βr 2α

1 Z αλr 4 0

πZ0 4 ω0 4 πZ0 ω0 βr 2α

1 Z αλr 2 0

πZ0 2 ω0 2 πZ0 ω0 βr 2α

Since (5.4.20) is similar to (5.1.17), this transmission line works as a series resonant circuit. Its equivalent-circuit parameters are found as follows (see Table 5.2): R ≈ Z0 αl = 14 Z0 αλr πZ0 4 ω0 4 C≈ πω0 Z0 L≈

(5.4.21) (5.4.22) (5.4.23)

and Q=

π β0 πZ0 = = 4R 4αl 2α

(5.4.24)

Example 5.5 Design a half-wavelength-long coaxial line resonator that is shortcircuited at its ends. Its inner conductor radius is 0.455 mm and the inner radius of the outer conductor is 1.499 mm. The conductors are made of copper. Compare the Q value of an air-filled resonator to that of a Teflon-filled resonator operating at 5 GHz. The dielectric constant of Teflon is 2.08 and its loss tangent is 0.0004. SOLUTION From the relations for coaxial lines given in Appendix 2, ωµ0 2π × 5 × 109 × 4π × 10−7 Rs = = = 0.018427 2σ 2 × 5.813 × 107 Rs 1 1 αc = + √ b 2 ln(b/a) µ0 /ε0 a

175

TRANSMISSION LINE RESONANT CIRCUITS

0.018427 = 2 × 376.7343 × ln(1.499/0.455)

100 100 + 0.455 1.499

= 0.058768 Np/m With Teflon filling, Rs αc = √ 2 µ0 /ε0 εr ln(b/a)

1 1 + a b

√ 100 100 0.018427 × 2.08 + = 2 × 376.7343 × ln(1.499/0.455) 0.455 1.499

= 0.084757 Np/m √ ω√ µ0 ε0 εr tan δ = π × 5 × 109 × 3 × 108 × 2.08 × 0.0004 αd = 2 = 0.030206 Np/m 2π × 5 × 109 = 104.719755 rad/m 3 × 108 √ 2π × 5 × 109 × 2.08 = 151.029 rad/m βd = 3 × 108 β0 104.719755 Qair = = = 890.96 2α 2 × 0.058768 βd 151.029 QTeflon = = = 656.8629 2α 2 × (0.084757 + 0.030206) β0 =

Example 5.6 Design a half-wavelength-long microstrip resonator using a 50- line that is short circuited at its ends. The substrate thickness is 0.159 cm, with its dielectric constant 2.08 and the loss tangent 0.0004. The conductors are made of copper. Compute the length of the line for resonance at 2 GHz and the Q of the resonator. Assume that thickness t of the microstrip is 0.159 µm. SOLUTION Design equations for the microstrip line (from Appendix 2) are 8eA e2A − 2 εr − 1 w B − 1 − ln(2B − 1) + = 2 2εr h 0.61 π × ln(B − 1) + 0.39 − εr where Z0 A= 60

εr + 1 2

1/2

for A > 1.52

for A ≤ 1.52

εr − 1 0.11 + 0.23 + εr + 1 εr

176

RESONANT CIRCUITS

w

h

Figure 5.20 Geometry of the microstrip line.

and B=

60π2 √ Z0 εr

See Figure 5.20. Therefore, 50 A= 60

2.08 + 1 2

1/2

2.08 − 1 0.11 + 0.23 + = 1.1333 2.08 + 1 2.08

Since A is smaller than 1.52, we need B to determine the width of microstrip. Hence, B=

60π2 = 8.212 √ 50 × 2.08

w 2 2.08 − 1 = 8.212 − 1 − ln(2 × 8.212 − 1) + h π 2 × 2.08 0.61 × ln(8.212 − 1) + 0.39 − = 3.1921 2.08 Therefore,

w = 0.507543 cm ≈ 0.51 cm

Further, εre = 0.5 2.08 + 1 + (2.08 − 1) × √ −

1 1 + 12/3.192094

(2.08 − 1)0.0001 √ = 1.7875 4.6 3.1921

Therefore, the length of the resonator can be determined as follows: l=

vp /f c 3 × 1010 λ = = √ cm = 5.6096 cm √ = 2 2 2f εe 2 × 2 × 109 × 1.7875

177

MICROWAVE RESONATORS

Since σcopper = 5.813 × 107 S/m, the quality factor of this resonator is determined as follows. For w/ h ≥ 1/2π, w t h we = + 0.3979 1 + ln 2 = 3.1923 h h h t and

1.25t 1.25 h h 1− + ln 2 = 2.5476 ζ=1+ we πh π t

Therefore, for w/ h ≥ 1, αc is found to be fGHz we 0.667 (we / h) −5 ζZ0 εre αc = 44.1255 × 10 + h σ h we / h + 1.444

Np/m

= 0.0428 Np/m Similarly, αd is found from (A2.28) as follows: αd = 10.4766

εr εre − 1 fGHz tan δ √ εr − 1 εre

Np/m = 0.0095 Np/m

and β=

2π 2π 2π = = rad/m = 56.0035 rad/m λ 2l 2 × 5.6096/100

Hence, Q of the resonator is found as Q=

56.0035 β = = 535.4 2α 2 × (0.0428 + 0.0095)

5.5 MICROWAVE RESONATORS Cavities and dielectric resonators are commonly employed as resonant circuits at frequencies above 1 GHz. These resonators provide much higher Q than the lumped elements and transmission line circuits. However, the characterization of these devices requires analysis of associated electromagnetic fields. Characteristic relations of selected microwave resonators are summarized in this section. Interested readers can find several excellent references and textbooks analyzing these and other resonators. Rectangular Cavities Consider a rectangular cavity made of conducting walls with dimensions a × b × c, as shown in Figure 5.21. It is filled with a dielectric material of dielectric constant εr and relative permeability µr . In general, it can support both TEmnp

178

RESONANT CIRCUITS

y c

b

x a

z

Figure 5.21 Geometry of the rectangular cavity resonator.

and TMmnp modes, which may be degenerate. The cutoff frequency of TEmn and TMmn modes can be determined from the formula n 2 m 2 300 fc (MHz) = √ + MHz µr εr 2a 2b The resonant frequency fr of a rectangular cavity operating in either TEmnp or TMmnp mode can be determined as follows: 300 fr (MHz) = √ µr εr

m 2 2a

n 2 p 2 + + 2b 2c

MHz

(5.5.1)

If the cavity is made of a perfect conductor and filled with a perfect dielectric, it will have infinite Q. However, it is not possible in practice. In the case of cavity walls having a finite conductivity (instead of infinite for the perfect conductor) but filled with a perfect dielectric (no dielectric loss), its quality factor Qc for TE10p mode is given by Qc |10p

√ 60b(acω µε)3 µr = 2 3 3 2 3 3 πRs (2p a b + 2bc + p a c + ac ) εr

(5.5.2)

The permittivity and permeability of the dielectric filling are given by ε and µ, respectively, ω is angular frequency, and Rs is the surface resistivity of walls, which is related to the skin depth δs and the conductivity σ as follows. 1 = Rs = σδs

ωµ 2σ

(5.5.3)

On the other hand, if the cavity is filled with a dielectric with its loss tangent as tan δ while its walls are made of a perfect conductor, the quality factor Qd is given as 1 Qd = (5.5.4) tan δ

179

MICROWAVE RESONATORS

When there is power loss in both the cavity walls and the dielectric filling, the quality factor Q is found from Qc and Qd as follows: 1 1 1 + = Q Qc Qd

(5.5.5)

Example 5.7 An air-filled rectangular cavity is made from a piece of copper WR-90 waveguide. If it resonates at 9.379 GHz in TE101 mode, find the required length c and the Q value of this resonator. SOLUTION Specifications of WR-90 are given in Table A3.3 as follows: a = 0.9 in. = 2.286 cm and b = 0.4 in. = 1.016 cm From (5.5.1), 1 = 2c

9379 300

2 −

1 2a

2 = 22.3383 ⇒ c = 2.238 cm

Next, the surface resistance, Rs , is determined from (5.5.3) as 0.0253 , and the Qc is found from (5.5.2) to be about 7858. Example 5.8 A rectangular cavity made of copper has inner dimensions a = 1.6 cm, b = 0.71 cm, and c = 1.56 cm. It is filled with Teflon (εr = 2.05 and tan δ = 2.9268 × 10−4 ). Find the TE101 mode resonant frequency and Q value of this cavity. SOLUTION From (5.5.1), 300 fr = √ 2.05

1 2 × 0.016

2 +

1 2 × 0.0156

2 ≈ 9379 MHz = 9.379 GHz

Since power dissipates both in the dielectric filling as well as in the sidewalls, overall Q will be determined from (5.5.5). From (5.5.2) and (5.5.4), Qc and Qd are found to be 5489 and 3417, respectively. Substituting these into (5.5.5), the Q value of this cavity is found to be 2106. Circular Cylindrical Cavities Figure 5.22 shows the geometry of a circular cylindrical cavity of radius r and height h. It is filled with a dielectric material of relative permeability µr and

180

RESONANT CIRCUITS z r

h y

x

Figure 5.22 Geometry of the circular cylindrical cavity resonator.

dielectric constant εr . Its resonant frequency fr in megahertz is given as χnm 2 pπ 2 300 fr (MHz) = + MHz (5.5.6) √ 2π µr εr r h

where χnm =

pnm p nm

for TM modes for TE modes

(5.5.7)

As discussed in Section 4.9, pnm represents the mth zero of the Bessel function of the first kind and order n and p nm represents the mth zero of the derivative of the Bessel function. In other words, these represent roots of the following equations, respectively (see Table 5.3): Jn (x) = 0

(5.5.8)

dJn (x) =0 dx

(5.5.9)

and

The Qc of a circular cylindrical cavity operating in TEnmp mode and filled with a lossless dielectric can be found from the following formula: 2 2 1.5 1 − n/pnm (pnm ) + (pπr/ h)2 47.7465 Qc = δs fr (MHz) p 2 + (2r/ h) (pπr/ h)2 + 1 − (2r/ h) npπr/p h2 nm

nm

(5.5.10)

TABLE 5.3 Zeros of Jn (x ) and Jn (x )

n

pn1

pn2

pn3

pn1

pn2

pn3

0 1 2

2.405 3.832 5.136

5.520 7.016 8.417

8.654 10.173 11.620

3.832 1.841 3.054

7.016 5.331 6.706

10.173 8.536 9.969

181

MICROWAVE RESONATORS

In the case of TMnmp mode, Qc is given by 47.7465 Qc = δs fr (MHz) For p = 0, Qc ≈

2 + (pπr/ h)2 pnm 1 + (2r/ h)

47.7465 pnm δs fr (MHz) 1 + (r/ h)

p>0

p=0

(5.5.11)

(5.5.12)

The quality factor Qd due to dielectric loss and the overall Q are determined as before from (5.5.4) and (5.5.5) with appropriate Qc . Example 5.9 Determine the dimensions of an air-filled circular cylindrical cavity that resonates at 5 GHz in TE011 mode. It should be made of copper, and its height should be equal to its diameter. Find the Q of this cavity given that n = 0, m = p = 1, and h = 2r. SOLUTION From (5.5.6), with χ01 as 3.832 from Table 5.3, we get 3.8322 + (π/2)2 r= = 0.03955 m (5000 × 2π/300)2 and h = 2r = 0.0791 m Since δs =

2 = ωµσ

2π × 5 ×

109

2 = 9.3459 × 10−7 m × 4π × 10−7 × 5.8 × 107

The Q value of the cavity can be found from (5.5.10) as Qc =

47.7465 × 106 [3.8322 + (π/2)2 ]1.5 × = 39, 984.6 9.3459 × 10−7 × 5 × 109 [3.8322 + (π/2)2 ] + 1

Further, there is no loss of power in air (tan δ ≈ 0), Qd is infinite, and therefore Q is the same as Qc . Dielectric Resonators Dielectric resonators (DRs), made of high-permittivity ceramics, provide high Q values with smaller size. These resonators are generally a solid sphere or cylinder of circular or rectangular cross section. High-purity TiO2 (εr ≈ 100, tan δ ≈ 0.0001) was used in early dielectric resonators. However, it was found

182

RESONANT CIRCUITS

h

2r

Figure 5.23 Geometry of an isolated circular cylindrical dielectric resonator.

to be intolerably dependent on the temperature. With the development of new ceramics, temperature dependence of DR can be reduced significantly. Figure 5.23 shows an isolated circular cylindrical DR. Its analysis is beyond the scope of this book. Its TE01δ -mode resonant frequency, fr in GHz, can be found from the following formula: fr (GHz) =

r 34 3.45 + rεr h

GHz

(5.5.13)

where r and h are in millimeters. This relation is found to be accurate within ±2% if r > 0.5 h

(5.5.14)

50 > εr > 30

(5.5.15)

2> and

An isolated DR is of almost no use in practice. A more practical situation is illustrated in Figure 5.24, where it is coupled with a microstrip line in a conducting enclosure. Assume that the dielectric constants of the DR and the substrate are εr and εs , respectively. With various dimensions as shown (all in meters), the following formulas can be used to determine its radius r and the height h. If r is selected between the following bounds, the formulas presented in this section are found to have a tolerance of no more than 2% (Kajfez and Guillon, 1986): 1.2892 × 108 1.2892 × 108 >r> (5.5.16) √ √ fr εs fr εr where fr is resonant frequency in hertz. The height of the DR is then found as α1 α2 1 −1 −1 tan + tan h= β β tanh α1 t β tanh α2 d

(5.5.17)

183

MICROWAVE RESONATORS

where k 2 − k02 εs α2 = k 2 − k02 β = k02 εr − k 2 α1 =

(5.5.18) (5.5.19) (5.5.20)

y 2.405 + r 2.405r[1 + (2.43/y) + 0.291y] y = (k0 r)2 (εr − 1) − 5.784

k =

and

√ k0 = ω µ0 ε0

(5.5.21) (5.5.22)

(5.5.23)

Example 5.10 Design a TE01δ -mode cylindrical dielectric resonator for use at 35 GHz in the microstrip circuit shown in Figure 5.24. The substrate is 0.25 mm thick and its dielectric constant is 9.9. The dielectric constant of the material available for DR is 36. Further, the top of the DR should have a clearance of 1 mm from the conducting enclosure. SOLUTION Since 1.2892 × 108 = 1.171 × 10−3 m √ fr εs and 1.2892 × 108 = 6.139 × 10−4 m √ fr εr the radius r of the DR may be selected as 0.835 mm. Conducting enclosure d Microstrip

DR

h Substrate

2r

t

Figure 5.24 Geometry of a circular cylindrical dielectric resonator in MIC configuration.

184

RESONANT CIRCUITS

The height h of the DR is then determined from (5.5.17) to (5.5.22), as follows: y = (k0 r)2 (εr − 1) − 5.784 = 2.707 2.405 y + = 3.382 × 103 r 2.405r[1 + (2.43/y) + 0.291y] α1 = k 2 − k02 εs = 2.474 × 103 α2 = k 2 − k02 = 3.302 × 103 β = k02 εr − k 2 = 2.812 × 103

k =

and

After substituting these numbers into (4.5.17), h is found to be α1 α2 1 h= tan−1 + tan−1 = 6.683 × 10−4 m = 0.668 mm β β tanh α1 t β tanh α2 d

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Kajfez, D., and P. Guillon, Dielectric Resonators. Dedham, MA: Artech House, 1986. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P.A., Microwave Engineering. Englewood Cliffs, NJ: Prentice Hall, 1988. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1998.

PROBLEMS 5.1. Determine the resonant frequency, bandwidth, and Q of the circuit shown in Figure P5.1. 2 nH

0.015 µF

vin(t)

50 Ω

Figure P5.1

vo(t)

185

PROBLEMS

5.2. The quality factor of a 100-nH inductor is 150 at 100 MHz. It is used in a series resonant circuit with a 50- load resistor. Find the capacitor required to resonate this circuit at 100 MHz. Find the loaded Q value of the circuit. 5.3. Find the capacitance C in the series R –L–C circuit shown in Figure P5.3 if it is resonating at 1200 rad/s. Compute the output voltage vo (t) when vin (t) = 4 cos(ωt + 0.2π) V for ω as 1200, 300, and 4800 rad/s. 10 mH

C

vin(t)

2Ω

vo(t)

Figure P5.3

5.4. The magnitude of input impedance in the series R –L–C circuit shown in Figure P5.4 is found to be 3 at a resonant frequency of 1600 rad/s. Find the inductance L, Q, and the bandwidth.

L

20 µF

vin(t)

vo(t) R

Figure P5.4

5.5. Determine the element values of a resonant circuit to filter all the sinusoidal signals from 100 to 130 MHz. This filter is to be connected between a voltage source with negligible internal impedance and a communication system that has an input impedance of 50 . Plot its response over the frequency range 50 to 200 MHz. 5.6. A series resonant circuit is resonating at 105 rad/s and has a bandwidth of 100 rad/s. If the resistance in the circuit is 50 , find its inductance and capacitance. 5.7. For the circuit given in Figure P5.7, the poles are located as illustrated in Figure P7.2. Find (a) the Q, (b) ω0 , (c) ω, (d) ω1 , and (e) ω2 .

186

RESONANT CIRCUITS

jω

L

C 98,000 −σ

vin(t)

50 Ω

−98,000

vo(t)

−2 × 104 (a)

(b)

Figure P5.7 (a) Series R –L–C circuit with input–output terminals; (b) poles of the transfer function.

5.8. Calculate the resonant frequency, Q, and bandwidth of the parallel resonant circuit shown in Figure P5.8.

2 nH

0.015 µF

100 kΩ

Figure P5.8

5.9. Find the resonant frequency, unloaded Q, and loaded Q of the parallel resonant circuit shown in Figure P5.9.

500 Ω 50 pF

20 nH Resonator circuit

1 kΩ

Load

Figure P5.9

5.10. The parallel R –L–C circuit shown in Figure P5.10 is resonant at 20,000 rad/s. If its admittance has a magnitude of 1 mS at the resonance, find the values of R and L.

187

PROBLEMS

L

0.05 µF

R

Figure P5.10

5.11. Consider the loaded parallel resonant circuit shown in Figure P5.11. Compute the resonant frequency in radians per second, unloaded Q, and loaded Q of this circuit.

50 µF

Load 20 mH

2 kΩ

10 kΩ

Figure P5.11

5.12. Resistance R in a parallel R –L–C circuit is 200 . The circuit has a bandwidth of 80 rad/s with a lower half-power frequency at 800 rad/s. Find the inductance and capacitance of the circuit. 5.13. A parallel R –L–C circuit is resonant at 2 × 106 rad/s and has a bandwidth of 20,000 rad/s. If its impedance at resonance is 2000 , find the circuit parameters. 5.14. A resonator is fabricated from a 2λ-long transmission line that is shortcircuited at its ends. Find its Q. 5.15. A 3-cm-long 100- air-filled coaxial line is used to fabricate a resonator. It is short-circuited at one end while a capacitor is connected at the other end to obtain the resonance at 6 GHz. Find the value of this capacitor. 5.16. A 100- air-filled coaxial line of length l is used to fabricate a resonator. It is terminated by 10 −j 5000 at its ends. If the signal wavelength is 1 m, find the required length for the first resonance and the Q of this resonator. 5.17. Design a half-wavelength-long coaxial line resonator that is short-circuited at its ends. Its inner conductor radius is 0.455 mm and the inner radius of the outer conductor is 1.499 mm. The conductors are made of copper. Compare the Q value of an air-filled resonator to that of a Teflon-filled resonator operating at 800 MHz. The dielectric constant of Teflon is 2.08 and its loss tangent is 0.0004. 5.18. Design a half-wavelength-long microstrip resonator using a 50- line that is short-circuited at its ends. The substrate thickness is 0.12 cm with a

188

RESONANT CIRCUITS

dielectric constant of 2.08 and a loss tangent of 0.0004. The conductors are of copper. Compute the length of the line for resonance at 900 MHz and the Q value of the resonator. 5.19. The secondary side of a tightly coupled transformer is terminated by a 2-k load in parallel with capacitor C2 . A current source I = 10 cos(4 × 106 ) mA is connected on its primary side. The inductance L1 of its primary is 0.30 µH, the mutual inductance M = 3 µH, and the unloaded Q of the secondary coil is 70. Determine the value of C2 such that this circuit resonates at 4 × 106 rad/s. Find its input impedance and the output voltage at the resonant frequency. What is the circuit bandwidth? 5.20. A 16- resistor terminates the secondary side of a tightly coupled transformer. A 40-µF capacitor is connected in series with its primary coil, which has an inductance L1 at 100 mH. If the secondary coil has an inductance L2 at 400 mH, find the resonant frequency and the Q value of the circuit. 5.21. A 16- resistor terminates the secondary side of a tightly coupled transformer. A 30-µF capacitor is connected in series with its primary coil, which has an inductance of 25 mH. If the circuit Q value is 50, find the inductance L2 of the secondary coil. 5.22. An air-filled rectangular cavity is made from a piece of copper WR-430 waveguide. If it resonates at 2 GHz in TE101 mode, find the required length c and the Q of this resonator. 5.23. Design an air-filled circular cylindrical cavity that resonates at 9 GHz in TE011 mode. It should be made of copper, and its height should be equal to its diameter. Find the Q value of this cavity. 5.24. Design a TE01δ mode cylindrical dielectric resonator for use at 4.267 GHz in the microstrip circuit shown in Figure 5.24. The substrate is 0.7 mm thick and its dielectric constant is 9.6. The dielectric constant of the material available for the DR is 34.19. Further, the top of the DR should have a clearance of 0.72 mm from the conducting enclosure.

6 IMPEDANCE-MATCHING NETWORKS

One of the most critical requirements in the design of high-frequency electronic circuits is that the maximum possible signal energy is transferred at each point. In other words, the signal should propagate in a forward direction with a negligible echo (ideally, zero). Echo signal not only reduces the power available but also deteriorates the signal quality due to the presence of multiple reflections. As noted in Chapter 5, impedance can be transformed to a new value by adjusting the turns ratio of a transformer that couples it with the circuit. However, it has several limitations. In this chapter we present a few techniques to design other impedance-transforming networks. These circuits include transmission line stubs and resistive and reactive networks. Further, the techniques introduced are needed in active circuit design at RF and microwave frequencies. As shown in Figure 6.1, impedance-matching networks are employed at the input and output of an amplifier circuit. These networks may also be needed to perform other tasks, such as filtering the signal and blocking or passing the dc bias voltages. This chapter begins with a section on impedance-matching techniques that use a single reactive element or stub connected in series or in shunt. Theoretical principles behind the technique are explained, and the graphical procedure to design these circuits using a Smith chart is presented. Principles and procedures of double-stub matching are discussed in the following section. The chapter ends with sections on resistive and reactive L-section matching networks. Both analytical and graphical procedures to design these networks using ZY charts are included. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

189

190

Zs

Vs

IMPEDANCE-MATCHING NETWORKS

Impedancetransforming circuit

Figure 6.1

Transistor with dc bias circuit

Impedancetransforming circuit

Load ZL

Block diagram of an amplifier circuit.

6.1 SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS When a lossless transmission line is terminated by an impedance ZL , the magnitude of the reflection coefficient (and hence, the VSWR) on it remains constant, but its phase angle can be anywhere between +180◦ and −180◦ . As we have seen in Chapter 3, it represents a circle on a Smith chart, and a point on this circle represents the normalized load. As one moves away from the load, the impedance (or the admittance) value changes. This movement is clockwise on the VSWR circle. The real part of the normalized impedance (or normalized admittance) becomes unity at certain points on the line. Addition of a single reactive element or a transmission line stub at this point can eliminate the echo signal and reduce the VSWR to unity beyond this point. A finite-length transmission line with its other end an open or short circuit is called a stub and behaves like a reactive element, as explained in Chapter 3. In this section we discuss the procedure for determining the location on a lossless feeding line where a stub or a reactive element can be connected to eliminate the echo signal. Two different possibilities, a series or a shunt element, are considered. Mathematical equations as well as graphical methods are presented to design the circuits.

Shunt Stub or Reactive Element Consider a lossless transmission line of characteristic impedance Z0 that is terminated by a load admittance YL , as shown in Figure 6.2. Corresponding normalized input admittance at a point ds away from the load can be found from (3.2.6) as Y L + j tan βds Y in = (6.1.1) 1 + j Y L tan βds To obtain a matched condition at ds , the real part of the input admittance must be equal to the characteristic admittance of the line [i.e., the real part of (6.1.1) must be unity]. This requirement is used to determine ds . The parallel susceptance Bs is then connected at ds to cancel out the imaginary part of Yin .

191

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

ds

YL = GL + jBL

jBs

Figure 6.2

Transmission line with a shunt matching element.

Hence, ds =

1 tan−1 β

2 B L ± B L − A(1 − GL ) A

(6.1.2)

2

where A = GL (GL − 1) + B L . The imaginary part of the normalized input admittance at ds is found as follows: 2

B in =

(B L + tan βds )(1 − B L tan βds ) − GL tan βds (GL tan βds )2 + (1 − B L tan βds )2

(6.1.3)

The other requirement to obtain a matched condition is B s = −B in

(6.1.4)

Hence, a shunt inductor is needed at ds if the input admittance is found capacitive (i.e., Bin is positive). On the other hand, it will require a capacitor if Yin is inductive at ds . As mentioned earlier, a lossless transmission line section can be used in place of this inductor or capacitor. The length of this transmission line section is determined according to the susceptance needed by (6.1.4) and the termination (i.e., an open circuit or a short circuit) at its other end. This transmission line section is called a stub. If ls is the stub length that has a short circuit at its other end, ls =

1 1 cot−1 (−B s ) = cot−1 B in β β

(6.1.5)

On the other hand, if there is an open circuit at the other end of the stub, ls =

1 1 tan−1 B s = tan−1 (−B in ) β β

(6.1.6)

Series Stub or Reactive Element If a reactive element (or stub) needs to be connected in series as shown in Figure 6.3, the design procedure can be developed as follows. The normalized

192

IMPEDANCE-MATCHING NETWORKS

ds

jXs ZL = RL + jXL

Figure 6.3 Transmission line with a matching element connected in series.

input impedance at ds is Z in =

Z L + j tan βds

(6.1.7)

1 + j Z L tan βds

To obtain a matched condition at ds , the real part of the input impedance must be equal to the characteristic impedance of the line [i.e., the real part of (6.1.7) must be unity]. This condition is used to determine ds . A reactance Xs is then connected in series at ds to cancel out the imaginary part of Zin . Hence,

ds =

1 tan−1 β

XL ±

2 X L − Az (1 − R L ) Az

(6.1.8)

2

where Az = R L (R L − 1) + XL . The imaginary part of the normalized input impedance at ds is found as follows: 2

X in =

(X L + tan βds )(1 − X L tan βds ) − R L tan βds (R L tan βds )2 + (1 − X L tan βds )2

(6.1.9)

To obtain a matched condition at ds , the reactive part Xin must be eliminated by adding an element of opposite nature. Hence, Xs = −Xin

(6.1.10)

Therefore, a capacitor will be needed in series if the input impedance is inductive. It will require an inductor if the input reactance is capacitive. As before, a transmission line stub can be used instead of an inductor or a capacitor. The length of this stub with an open circuit at its other end can be determined as follows: ls =

1 1 cot(−X s ) = cot Xin β β

(6.1.11)

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

193

However, if the stub has a short circuit at its other end, its length will be a quarter-wavelength shorter (or longer, if the resulting number becomes negative) than this value. It can be found as ls =

1 1 tan Xs = tan(−X in ) β β

(6.1.12)

Note that the location ds and the stub length ls are periodic in nature in both cases. It means that the matching conditions will also be satisfied at points one-half wavelength apart. However, the shortest possible values of ds and ls are preferred because those provide the matched condition over a broader frequency band. Graphical Method These matching networks can also be designed graphically using a Smith chart. The procedure is similar for both series- and shunt-connected elements, except that the former is based on the normalized impedance, whereas the latter works with normalized admittance. It can be summarized in the following steps: 1. Determine the normalized impedance of the load and locate that point on a Smith chart. 2. Draw the constant-VSWR circle. If the stub needs to be connected in parallel, move a quarter-wavelength away from the load impedance point. This point is located at the other end of the diameter that connects the load point with the center of the circle. For a series stub, stay at the normalized impedance point. 3. From the point found in step 2, move toward the generator (i.e., clockwise) on the VSWR circle until it intersects the unity resistance (or conductance) circle. The distance traveled to this intersection point from the load is equal to ds . There will be at least two such points within one-half wavelength from the load. A matching element can be placed at either one of these points. 4. If the admittance in step 3 is 1 + j B, a susceptance of −j B in shunt is needed for matching. This can be a discrete reactive element (inductor or capacitor, depending on a negative or positive susceptance value) or a transmission line stub. 5. In the case of a stub, the length required is determined as follows. Since its other end will have an open or a short, the VSWR on it will be infinite. It is represented by the outermost circle of the Smith chart. Locate the desired susceptance point (i.e., 0 − j B) on this circle and then move toward load (counterclockwise) until an open circuit (i.e., a zero susceptance) or a short circuit (an infinite susceptance) is found. This distance is equal to the stub length ls . For a series reactive element or stub, steps 4 and 5 will be the same except that the normalized reactance replaces the normalized susceptance.

194

IMPEDANCE-MATCHING NETWORKS

Example 6.1 A uniform, lossless 100- line is connected to a load of 50 − j 75 , as illustrated in Figure 6.4. A single stub of 100- characteristic impedance is connected in parallel at a distance ds from the load. Find the shortest values of ds and stub length ls for a match. SOLUTION As mentioned in the preceding analysis, design equations (6.1.2), (6.1.3), (6.1.5), and (6.1.6) for a shunt stub use admittance parameters. On the other hand, the series-connected stub design uses impedance parameters in (6.1.8), (6.1.9), (6.1.11), and (6.1.12). Therefore, ds and ls can be determined theoretically, as follows: YL =

YL Z0 100 = = = 0.6154 + j 0.9231 Y0 ZL 50 − j 75 2

A = GL (GL − 1) + B L = 0.6154(0.6154 − 1) + 0.92312 = 0.6154 From (6.1.2), the two possible values of ds are λ (−0.75)2 − 0.6154(1 − 0.5) −1 −0.75 + ds = tan = 0.1949λ 2π 0.6154 and λ −0.75 − tan−1 ds = 2π

(−0.75)2 − 0.6154(1 − 0.5) = 0.0353λ 0.6154

At 0.1949λ from the load, the real part of the normalized admittance is unity and its imaginary part is −1.2748. Hence, the stub should provide j 1.2748 to cancel it out. Length of a short-circuited stub, ls , is calculated from (6.1.5) as ls =

1 cot−1 (−1.2748) = 0.3941λ β

ds

ZL = 50 − j75 Ω

ls

Figure 6.4 Shunt stub matching network.

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

195

On the other hand, normalized admittance is 1 + j 1.2748 at 0.0353λ from the load. To obtain a matched condition, the stub at this point must provide a normalized susceptance of −j 1.2748. Hence, ls =

1 cot−1 (1.2748) = 0.1059λ β

Thus, there are two possible solutions to this problem. In one case, a shortcircuited 0.3941λ-long stub is needed at 0.1949λ from the load. The other design requires a 0.1059λ-long short-circuited stub at 0.0353λ from the load. It is preferred over the former design because of its shorter lengths. The following steps are needed to solve this example graphically with a Smith chart: 1. Determine the normalized load admittance. ZL =

50 − j 75 = 0.5 − j 0.75 100

2. Locate the normalized load impedance point on the Smith chart. Draw the VSWR circle as shown in Figure 6.5. 3. From the load impedance point, move to the diametrically opposite point and locate the corresponding normalized load admittance. It is the point 0.62 + j 0.91 on the chart. 4. Locate the point on the VSWR circle where the real part of the admittance is unity. There are two such points with normalized admittance values 1 + j 1.3 (say, point A) and 1 − j 1.3 (say, point B), respectively. 5. The distance ds of 1 + j 1.3 (point A) from the load admittance can be determined as 0.036λ (i.e., 0.170λ − 0.134λ) and for point B (1 − j 1.3) as 0.195λ (i.e., 0.329λ − 0.134λ). 6. If a susceptance of −j 1.3 is added at point A or j 1.3 at point B, the load will be matched. 7. Locate the point −j 1.3 along the lower circumference of the chart and from there move toward the load (counterclockwise) until the short circuit (infinity on the chart) is reached. Separation between these two points is 0.25λ − 0.146λ = 0.104λ. Hence a 0.104λ-long transmission line with a short circuit at its rear end will have the desired susceptance for point A. 8. For a matching stub at point B, locate the point j 1.3 on the upper circumference of the chart and then move toward the load up to the short circuit (i.e., the right-hand end of the chart). Hence, the stub length ls for this case is determined as 0.25λ + 0.146λ = 0.396λ. Therefore, a 0.104λ-long stub at 0.036λ from the load (point A) or a 0.396λlong stub at 0.195λ (point B) from the load will match the load. These values are comparable to those obtained earlier.

196

1

0.8

0.15 0.35

2

0.4

0 0.2 0 0.3

S TOWARD GE N ENGTH VEL THS TOWARD LERATO -WA - AVELENG 0.0 OAD R -W - -> LEN 0.0 E V A W

-WA

4

0.

0 0.2 0 0.3

0.0 0.4 5 5

0.5

0.6

0.7

0.10 0.40

0.9

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

0.2

5

10 10

5

2

0.5

0.4

0.3

0.2

0.1

0.1

0.25 0.25

10

0.1

0.2

5 0.3 0.2 0 0

4

0.

1

0.9

0.8

0.7

0.40 0.10

Figure 6.10

0.35 0.15

0.6

2

0.5

5 0.4 5 0.0

0.3

Graphical solution to Example 6.3.

the corresponding input impedance point is identified as 1.27 + j 0.37. For the circuit given in Figure 6.9(a), the admittance (normalized) points are found as 0.45 + j 0.22 and 0.73 − j 0.21, respectively. Next, move from the normalized load admittance point toward the generator by a distance of 0.083583λ (i.e., 0.042λ + 0.084λ = 0.126λ of the scale “wavelengths toward generator”). The normalized admittance value of this point is found to be 0.73 + j 0.69. Hence, its real part is the same as that of the desired input admittance. However, its susceptance is j 0.69, whereas the desired value is −j 0.2. Hence, an inductor will be needed in parallel at this point. Since the given circuit has a capacitor, this design is not possible. For the circuit shown in Figure 6.9(b), elements are connected in series. Therefore, normalized impedance points need to be used in this case. Move from the load impedance (1.8 − j 0.85) point toward the generator by a distance of 0.029928λ (i.e., 0.292λ + 0.03λ = 0.322λ on the “wavelengths toward generator” scale). Normalized impedance value at this point is 1.27 − j 0.95. Thus,

202

IMPEDANCE-MATCHING NETWORKS

the resistance at this point is found to be equal to the value desired. However, its reactance is −j 0.95, whereas the required value is j 0.37. Therefore, a series reactance of j 1.32 is needed at this point. The given circuit has an inductor that provides a positive reactance. Hence, this circuit will work. The required inductance L is found as L=

1.32 × 50 H = 2.626 nH 2π × 4 × 109

6.2 DOUBLE-STUB MATCHING NETWORKS The matching technique presented in Section 5.1 requires that a reactive element or stub be placed at a precise distance from the load. This point will shift with load impedance. Sometimes it may not be feasible to match the load using a single reactive element. Another possible technique to match the circuit employs two stubs with a fixed separation between them. This device can be inserted at a convenient point before the load. The impedance is matched by adjusting the lengths of the two stubs. Of course, it does not provide a universal solution. As will be seen later in this section, separation between the two stubs limits the range of load impedance that can be matched with a given double-stub tuner. Let l1 and l2 be the lengths of two stubs, as shown in Figure 6.11. The first stub is located at a distance l from the load, ZL = RL + j XL ohms. Separation between the two stubs is d, and the characteristic impedance of every transmission line is Z0 . In double-stub matching, the load impedance ZL is transformed to the normalized admittance at the location of the first stub. Since the stub is connected in parallel, its normalized susceptance is added to that and the resulting normalized admittance is transferred to the location of the second stub. Matching conditions at this point require that the real part of this normalized admittance

l

d

Z0

Z0

l2

Figure 6.11

ZL = RL + jXL

l1

Double-stub matching network.

DOUBLE-STUB MATCHING NETWORKS

203

be equal to unity while its imaginary part is canceled by a conjugate susceptance of the second stub. Mathematically, Y + j (B 1 + tan βd) 1 + j (Y + j B 1 ) tan βd where Y =

1 + j Z L tan βl Z L + j tan βl

=

+ j B2 = 1

Y L + j tan βl 1 + j Y L tan βl

= G + jB

(6.2.1)

(6.2.2)

j B1 and j B2 are the susceptance of the first and second stubs, respectively, and β is the propagation constant over the line. For Re

Y + j (B 1 + tan βd) 1 + j (Y + j B 1 ) tan βd

=1

2

G tan2 βd − G(1 + tan2 βd) + [1 − (B + B 1 ) tan βd]2 = 0

(6.2.3)

Since conductance of the passive network must be a positive quantity, (6.2.3) requires that a given double stub can be used for matching only if the following condition is satisfied: 0 ≤ G ≤ csc2 βd (6.2.4) Two possible susceptances of the first stub that can match the load are determined by solving (6.2.3):

2 2 B 1 = cot βd 1 − B tan βd ± G sec βd − (G tan βd)

(6.2.5)

Normalized susceptance of the second stub is determined from (6.2.1) as 2

B2 =

G tan βd − (B + B 1 + tan βd)[1 − (B + B 1 ) tan βd] (G tan βd)2 + [1 − (B + B 1 ) tan βd]2

(6.2.6)

Once the susceptance of a stub is known, its short-circuit length can be determined easily as follows: 1 l1 = cot−1 (−B 1 ) (6.2.7) β and l2 =

1 cot−1 (−B 2 ) β

(6.2.8)

204

IMPEDANCE-MATCHING NETWORKS

Graphical Method A two-stub matching network can also be designed graphically with the help of a Smith chart. This procedure follows the analytical concepts discussed above and can be summarized as follows: 1. Locate the normalized load-impedance point on a Smith chart and draw the VSWR circle. Move to the corresponding normalized admittance point. If the load is connected right at the first stub, go to the next step; otherwise, move toward the generator (clockwise) by 2βl on the VSWR circle. Assume that the normalized admittance of this point is g + j b. 2. Rotate the unity-conductance circle counterclockwise by 2βd. The conductance circle that touches this circle encloses the “forbidden region.” In other words, this tuner can match only those Y that lie outside this circle. It is a graphical representation of the condition expressed by (6.2.4). 3. From g + j b, move on the constant-conductance circle until it intersects the rotated unity-conductance circle. There are at least two such points, providing two design possibilities. Let the normalized admittance of one of these points be g + j b1 . 4. The normalized susceptance required of the first stub is j (b1 − b). 5. Draw a VSWR circle through point g + j b1 and move toward the generator (clockwise) by 2βd on it. This point will fall on the unity-conductance circle of the Smith chart. Assume that this point is 1 + j b2 . 6. The susceptance required from the second stub is −j b2 . 7. Once the stub susceptances are known, their lengths can be determined following the procedure used in the technique described earlier. Example 6.4 For the double-stub tuner shown in Figure 6.12, find the shortest values of l1 and l2 to match the load.

λ/8

λ/8 Z0 = 50 Ω

Z0 = 50 Ω

Z0 = 50 Ω

ZL = 100 + j50 Ω Z0 = 50 Ω

l2

l1

Figure 6.12 Two-stub matching network for Example 6.4.

205

DOUBLE-STUB MATCHING NETWORKS

SOLUTION Since the two stubs are separated by λ/8, βd is equal to π/4 and the condition (6.2.4) gives 0≤G≤2 This means that the real part of the normalized admittance at the first stub (loadside stub) must be less than 2; otherwise, it cannot be matched. The graphical procedure requires the following steps to find stub settings: 100 + j 50 = 2 + j1 50 2. Locate the normalized load impedance on a Smith chart (point A in Figure 6.13) and draw the VSWR circle. Move to the diametrically opposite side and locate the corresponding normalized admittance point B at 0.4 − j 0.2.

0.8

1

0.15 0.35

2

F 4

0.

0 0.2 0 0.3

0.0 0.4 5 5

0.5

0.6

0.7

0.10 0.40

0.9

1. Z L =

0.2

5

E A

5

2

10

0.5

0.4

0.3

0.2

0.1

0.1

D

10 0.25 0.25

OWARD GENERA GTHS T TO LEN E S TOWARD LOAD R ---> V H T -WA VELENG A 0.0 -W 10 ), Rp is 50 while Rs is 10 . Therefore, Q2 =

50 −1=4⇒Q=2 10

214

IMPEDANCE-MATCHING NETWORKS

Xs and Xp can now be determined from (6.3.8) and (6.3.9): Xs = QRs = 2 × 10 = 20 and Xp =

50 Rp = = 25 Q 2

Reactance Xp is connected in parallel with the transmission line and Xs is in series with the load. These two matching circuits are shown in Figure 6.20. If Xp is a capacitive reactance of 25 , Xs must be inductive 20 . The reactive part that has not been taken into account so far needs to be included at this point in Xs . Since this reactive part of the load is inductive 10 , another series inductor for the remaining 10 is needed, as shown in circuit (1) of Figure 6.20. Hence, ωLs = 20 − 10 Therefore, Ls =

10 H = 0.3183 × 10−8 H = 3.183 nH 2π × 500 × 106

and

1 = 25 ωCp

Therefore, Cp =

1 = 0.0127324 × 10−9 F = 12.7324 pF 2π × 500 × 106 × 25

In the second case, Xp is assumed inductive and therefore Xs needs to be capacitive. It is circuit (2) in Figure 6.20. Since the load has a 10- inductive reactance,

Cs

Ls

10 Ω Cp 50 Ω

Load

10 Ω 50 Ω

Lp

j10 Ω

j10 Ω (1)

Load

(2)

Figure 6.20 Reactive matching circuits for Example 6.8.

MATCHING NETWORKS USING LUMPED ELEMENTS

215

it must be taken into account in determining the required capacitance. Therefore, ωCs =

1 1 ⇒ Cs = F = 10.61 pF 20 + 10 2π × 500 × 106 × 30

and ωLp = 25 ⇒ Lp =

25 H = 7.9577 nH 2π × 500 × 106

The matching procedure used so far is based on the transformation of series impedance to a parallel circuit of the same quality factor. Similarly, one can use an admittance transformation, as shown in Figure 6.21. These design equations can be obtained easily following the procedure used earlier: Gs = 1 + Q2 Gp

(6.3.11)

Q=

Gs Bs

(6.3.12)

Q=

Bp Gp

(6.3.13)

and

Note from (6.3.11) that Gs must be greater than Gp for a real Q. Example 6.9 A load admittance, YL = 8 − j 12 mS, is to be matched with a 50- line. There are three different L-section matching networks given to you, as shown in Figure 6.22. Complete or verify each of these circuits. Find the element values at 1 GHz. SOLUTION Since the characteristic impedance of the line is 50 , the corresponding conductance will be 0.02 S. The real part of the load admittance is 0.008 S. Therefore, Gs must be 0.02 S in (6.3.11). However, the circuits in Figure 6.22 have a capacitor or an inductor connected in parallel with the 50- line. Obviously, these design equations cannot be used here. Equations (6.3.8)

Gs

Gp jBs

jBp

Figure 6.21 Series- and parallel-connected admittance circuits.

216

IMPEDANCE-MATCHING NETWORKS Cs

Ls

Load

Load Cp 50 Ω

Lp

50 Ω

YL

(a)

YL

(b) Cs

Load

Cp 50 Ω

YL

(c)

Figure 6.22 Three possible L-section matching circuits given in Example 6.9.

to (6.3.10) use the impedance values. Hence, we follow the procedure outlined below: YL =

1 103 ⇒ ZL = = 38.4615 + j 57.6923 ZL 8 − j 12

Therefore, Rp = 50 and Rs = 38.4615 ⇒ Q =

50 − 1 = 0.5477 38.4615

Now, from (6.3.8) and (6.3.9), Xs = QRs = 0.5477 × 38.4615 = 21.0654 and Xp =

50 Rp = = 91.2909 Q 0.5477

For circuit (a) in Figure 6.22, capacitor Cp can be selected, which provides 91.29 reactance. It means that the inductive reactance on its right-hand side must be 21.06 . However, an inductive reactance of 57.69 is already present there, due to load. Another series inductor will increase it further, whereas it needs to be reduced to 21.06 . Thus, we conclude that circuit (a) cannot be used.

217

MATCHING NETWORKS USING LUMPED ELEMENTS

In circuit (b) in Figure 6.22, there is an inductor in parallel with the transmission line. For a match, its reactance must be 91.29 , and overall reactance to the right must be capacitive 21.06 . Inductive reactance of 57.69 of the load needs to be neutralized as well. Hence, Xc = 57.6923 + 21.0654 = 78.7577 Therefore, C=

2π ×

109

1 F = 2.0208 pF × 78.7577

and Xp = 91.2909 ⇒ L =

91.2909 H = 14.5294 nH 2π × 109

Circuit (c) in Figure 6.22 has a capacitor across the transmission line terminals. Assuming that its reactance is 91.29 , reactance to the right must be inductive 21.06 . As noted before, the reactive part of the load is an inductive 57.69 . It can be reduced to the desired value by connecting a capacitor in series. Hence, the component values for circuit (c) are calculated as follows: Xs = 57.6923 − 21.0654 = 36.6269 Therefore, Cs = 4.3453 pF and Xp = 91.2909 ⇒ Cp = 1.7434 pF Example 6.10 Reconsider Example 6.9, where YL = 8 −j 12 mS is to be matched with a 50- line. Complete or verify the L-section matching circuits shown in Figure 6.23 at a signal frequency of 1 GHz.

Ls Cs

Cp 50 Ω

YL

(a)

Figure 6.23

Load

Load 50 Ω

Cp

YL

(b)

L-section matching networks given in Example 6.10.

218

IMPEDANCE-MATCHING NETWORKS

SOLUTION As noted in Example 6.9, Gs is 0.02 S and Gp is 8 mS. Looking over the given circuit configurations, it seems possible to complete or verify these circuits through (6.3.11) to (6.1.13). Hence, Gs 0.02 Q= − 1 = 1.2247 −1= Gp 0.008 Bs =

Gs 0.02 = = 0.0163 S Q 1.2247

and Bp = QGp = 1.2247 × 0.008 = 0.0097978 ≈ 0.01 S If Bs is selected as an inductive susceptance, Bp must be capacitive. Since load has an inductive susceptance of 12 mS, the capacitor must neutralize that, too. Hence, circuit (a) of Figure 6.23 will work provided that Bs = 0.0163 S ⇒ Ls = 9.746 nH and Bp = (0.01 + 0.012) S = 0.022 S ⇒ Cp = 3.501 pF In circuit (b) of Figure 6.23, a capacitor is connected in series with Gs . Therefore, Bp must be an inductive susceptance. Since it is 0.012 S whereas only 0.01 S is required for matching, a capacitor in parallel is needed. Hence, Bs = 0.0163 S ⇒ Cs = 2.599 pF and Bp = (0.012 − 0.01) = 0.002 S (capacitive) ⇒ Cp = 0.3183 pF Thus, both circuits can work provided that the component values are as found above. Graphical Method As described earlier, L-section reactive networks can be used for matching the impedance. One of these reactive elements appears in series with the load or the desired impedance, while the other one is connected in parallel. Thus, the resistive part stays constant when a reactance is connected in series with impedance. Similarly, a change in the shunt-connected susceptance does not affect the conductive part of the admittance. Consider the normalized impedance point X on the Smith chart shown in Figure 6.24. To distinguish it from others, it may be called the Z-Smith chart. If its resistive part needs to be kept constant at 0.5, one must stay on this constant-resistance circle. A clockwise movement from

219

MATCHING NETWORKS USING LUMPED ELEMENTS

0

jX

12

110

90

100

80

70 60 50

30

40

14 0

1

Inductive reactance

15 0

Capacitive reactance 170

10

0 10

5

2

1

0.5

0.2

−170

−16

−20

0

−10

Capacitive reactance

0

180

Inductive reactance

20

160

30

X

−

0 15

−3 0

−

0

0 14

−4 −5

0

30

−6

0

−70

20

−80

−1

0 − 1

−90

−100 −11

-jX

Figure 6.24 Impedance (Z-) Smith chart.

this point increases the positive reactance. It means that the inductance increases in series with the impedance of X. On the other hand, the positive reactance decreases with a counterclockwise movement. A reduction in positive reactance means a decrease in series inductance or an increase in series capacitance. Note that this movement also represents an increase in negative reactance. Now consider a Smith chart that is rotated by 180◦ from its usual position, as shown in Figure 6.25. It may be called a Y -Smith chart because it represents the admittance plots. In this case, addition (or subtraction) of a susceptance to admittance does not affect its real part. Hence, it represents a movement on the constant-conductance circle. Assume that a normalized admittance is located at point X. The conductive part of this admittance is 0.5. If a shunt inductance is added, it moves counterclockwise on this circle. On the other hand, a capacitive susceptance moves this point clockwise on the constant-conductance circle. A superposition of Z- and Y -Smith charts is shown in Figure 6.26. It is generally referred to as a ZY -Smith chart because it includes impedance as well as admittance plots at the same time. A short-circuit impedance is zero while the corresponding admittance goes to infinity. A single point on the ZY -Smith chart represents it. Similarly, it may be found that other impedance points of the Z chart coincide with the corresponding admittance points of the Y chart as well. Hence, impedance

220

IMPEDANCE-MATCHING NETWORKS

12 0

−100

−90

−80

−70

−6

0

−

−5

0

−1

4 −1

0

30

-110

−4 0

-jB

X

−10

0

0.2

0.5

1

2

5

0 10

Inductive susceptance

180

Capacitive susceptance

−170

−20

1 0 − −16

50

−3 0

Inductive susceptance

170

10

20

160

Capacitive susceptance

15

0

30 14

0

40 50

0

13

60

70

20

80

Figure 6.25

90

100

110

jB

1

Admittance (Y -) Smith chart..

can be transformed to corresponding admittance just by switching from the Z- to the Y -scales of a ZY -Smith chart. For example, a normalized impedance point of 0.9 − j 1 is located using the Z-chart scales as A in Figure 6.26. The corresponding admittance is read from a Y -chart as 0.5 + j 0.55. Reactive L-section matching circuits can be designed easily using a ZY -Smith chart. Load and desired impedance points are identified on this chart using Zscales. Y -scales may be used if either one is given as admittance. Note that the same characteristic impedance is used for normalizing these values. Now, move from the point to be transformed toward the desired point following a constantresistance or constant-conductance circle. Movement on the constant-conductance circle gives the required susceptance (i.e., a reactive element will be needed in shunt). It is determined by subtracting initial susceptance (the starting point) from the value reached. When moving on the conductance circle, use the Y -scale of the chart. Similarly, moving on a constant-resistance circle will mean that a reactance must be connected in series. It can be determined using the Z-scale of the ZY -Smith chart.

221

MATCHING NETWORKS USING LUMPED ELEMENTS

110

90

100

80

70 60

0 12

50

0

10

170

20

160

15

30

0

40

14 0

13

10

5

2

1

0.5

−20

0 −16

−10

−170

0.2

0

180

0

−

0

−4

0 14

0

−1

50

−3

A

−5

30

0

−6

0

20

−70

−80

−90

−100

0

−11

−1

−1

Figure 6.26 ZY -chart.

Note from the ZY -chart that if a normalized load value falls inside the unity resistance circle, it can be matched with the characteristic impedance Z0 only after shunting it with an inductor or capacitor (a series-connected inductor or capacitor at the load will not work). Similarly, a series inductor or capacitor is required at the load if the normalized value is inside the unity conductance circle. Outside these two unity circles, if the load point falls in the upper half, the first component required for matching would be a capacitor that can be series or parallel connected. On the other hand, an inductor will be needed at the load first if the load point is located in the lower half and outside the unity circles. Example 6.11 Use a ZY -Smith chart to design the matching circuits of Examples 6.9 and 6.10.

222

IMPEDANCE-MATCHING NETWORKS

SOLUTION In this case the given load admittance can be normalized by the characteristic admittance of the transmission line. Hence, YL =

YL = YL Z0 = 0.4 − j 0.6 Y0

This point is found on the ZY -Smith chart as A in Figure 6.27. The matching requires that this admittance must be transformed to 1. This point can be reached through a unity conductance or resistance circle. Hence, a matching circuit can be designed if somehow we can reach one of these circles through constantresistance and constant-conductance circles only. Since constant-conductance circles intersect constant- resistance circles, one needs to start on a constant-

110

90

100

80

70 60

0 12

50

0

14

40

0

13

0 15

30

A

20

160

B

10

170

D

10

5

2

1

0.5

O

−10

−170

0.2

0

180

0

−20

0 −16

E

−4 0

40 −1

−

0 15

−3 0

C

−5

30

0

−6

0

20

−70

−80

−90

−100

−11 0

−1

−1

Figure 6.27 Graphical solution to Example 6.11.

MATCHING NETWORKS USING LUMPED ELEMENTS

223

resistance circle to get on the unity conductance circle, or vice versa. There are two circles (i.e., 0.4 conductance and 0.78 resistance circles) passing through point A. Hence, it is possible to get on to the unity resistance circle via the 0.4 conductance circle. Alternatively, one can reach on the unity conductance circle through the 0.78 resistance circle. In either case, a circuit can be designed. From point A, one can move on the 0.4 conductance circle (or 0.78 resistance circle) so that the real part of the admittance (or impedance) remains constant. If we start counterclockwise from A on the conductance circle, we end up at infinite susceptance without intersecting the unity resistance circle. Obviously, this does not yield a matching circuit. Similarly, a clockwise movement on the 0.78 resistance circle from A cannot produce a matching network. Hence, an inductor in parallel or in series with the given load cannot be used for the design of a matching circuit. This proves that the circuit shown in Figure 6.22(a) cannot be designed. There are four possible circuits. In one case, move from point A to B on the conductance circle (a shunt capacitor) and then from B to O on the unity resistance circle (a series capacitor). The second possibility is via A to C (a shunt capacitor) and then from C to O (a series inductor). A third circuit can be obtained following the path from A to D (a series capacitor) and then from D to O (a shunt capacitor). The last circuit can be designed by following the path from A to E (a series capacitor) and then from E to O (a shunt inductor). A–E to E–O and A–D to D–O correspond to the circuits shown in Figure 6.22(b) and (c), respectively. Similarly, A–C to C–O and A–B to B–O correspond to circuits shown in Figure 6.23(a) and (b), respectively. Component values for each of these circuits are determined following the corresponding susceptance or reactance scales. For example, we move along a resistance circle from A to D. That means that there will be a change in reactance. The required element value is determined after subtracting the reactance at A from the reactance at D. This difference is j 0.42 − j 1.15 = −j 0.73. Negative reactance means that it is a series capacitor. Note that −j 0.73 is a normalized value. The actual reactance is −j 0.73 × 50 = −j 36.5 . It is close to the corresponding value of 36.6269 obtained earlier. Other components can be determined as well. Example 6.12 Two types of L-section matching networks are shown in Figure 6.28. Select one that can match the load ZL = 25 + j 10 to a 50- transmission line. Find the element values at 500 MHz. SOLUTION First, let us consider circuit (a). Rp must be 50 because it is required to be greater than Rs . However, the reactive element is connected in series with it. That will be possible only with Rs . Hence, this circuit cannot be designed using (6.3.8) to (6.3.10). The other set of design equations, namely, (6.3.11) to (6.3.13), requires admittance instead of impedance. Therefore, the given impedances are inverted to find the corresponding admittances as

224

IMPEDANCE-MATCHING NETWORKS

Ls

Cs

Z0 = 50 Ω

Lp

ZL

Z0 = 50 Ω

Cp

ZL

(b)

(a)

Figure 6.28 L-section matching circuits given for Example 6.12.

follows: YL =

1 1 = 0.034483 − j 0.01379 S = ZL 25 + j 10

Y0 =

1 1 = 0.02 S = Z0 50

and

Since Gs must be larger than Gp , this set of equations produces a matching circuit that has a reactance in series with YL . Thus, we conclude that the circuit given in Figure 6.28(a) cannot be designed. Now consider the circuit shown in Figure 6.28(b). For (6.3.8) to (6.3.10), Rp is 50 while Rs is 25 . Hence, one reactance of the matching circuit will be connected across 50 and the other one will go in series with ZL . The given circuit has this configuration. Hence, it will work. Its component values are calculated as 50 Q= −1=1 25 Therefore, Xs = Rs = 25 and Xp = Rp = 50 For Xp to be capacitive, we have 1 = 50 ⇒ Cp = 6.366 pF ωCp

225

MATCHING NETWORKS USING LUMPED ELEMENTS

As 10 of inductive reactance is already included in the load, another inductive 15 will suffice. Thus, ωLs = 25 − 10 = 15 ⇒ Ls =

15 H = 4.775 nH ω

This example can be solved using a ZY -Smith chart as well. To that end, the load impedance is normalized with characteristic impedance of the line. Hence, ZL =

25 + j 10 = 0.5 + j 0.2 50

This point is located on a ZY -Smith chart as shown in Figure 6.29. It is point A on this chart. If we move on the conductance circle from A, we never reach

90

100

110

80

70 60

0 12

50

0

15

30

0

14

40

0

13

10

160 170

20

B

A O 10

5

2

1

0.5

−20

0 −16

−10

−170

0.2

0

180

0

0 14

50 −1

−3 0

−

0

−4 −5

30

0

−6

0

20

−70

−80

−90

−100

−11 0

−1

−1

Figure 6.29 Graphical solution to Example 6.12.

226

IMPEDANCE-MATCHING NETWORKS

the unity-resistance circle. This means that an L-section matching circuit that has a reactance (inductor or capacitor) in parallel with the load cannot be designed. Therefore, circuit (a) of Figure 6.28 is not possible. Next, we try a reactance in series with the load. Moving on the constantresistance circle from A indicates that there are two possible circuits. Movement from A to B on the resistance circle and then from B to O on the unity conductance circle provides component values of circuit shown in Figure 6.28(b). The normalized reactance required is found to be j 0.3 (i.e., j 0.5 − j 0.2). It is an inductor because of its positive value. Further, it will be connected in series with the load because it is found by moving on a constant-resistance circle from A. The series inductance required is determined as XL = 0.3 × 50 = 15 ⇒ Ls =

15 H = 4.775 nH ω

Movement from B to O gives a normalized susceptance of j 1. Positive susceptance is a capacitor in parallel with a 50- line. Hence, ωCp =

1 ⇒ Cp = 6.366 pF 50

These results are found to be exactly equal to those found earlier analytically. SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P. A., Microwave Engineering, Englewood Cliffs, NJ: Prentice Hall, 1988. Sinnema, W., Electronic Transmission Technology. Englewood Cliffs, NJ: Prentice Hall, 1988.

PROBLEMS 6.1. A 10-V (rms) voltage source in series with a 50- resistance represents a signal generator. It is to be matched with a 100- load. Design a matching circuit that provides perfect matching over the frequency band 1 kHz to 1 GHz. Determine the power delivered to the load.

227

PROBLEMS

6.2. A 1000- load is to be matched with a signal generator that can be represented by a 1-A (rms) current source in parallel with a 100- resistance. Design a resistive circuit to match it. Determine the power dissipated in the matching network. 6.3. Design a single-stub network to match a 800 −j 300- load to a 400- lossless line. The stub should be located as close to the load as possible and it is to be connected in parallel with the transmission line. 6.4. A 140 −j 70- load is terminating a 70- transmission line, as shown in Figure P6.4. Find the location and length of a short-circuited stub of a 40- characteristic impedance that will match the load with the line.

70 Ω

70 Ω 140 − j70 Ω

40 Ω

S.C.

Figure P6.4

6.5. An antenna has impedance of 40 +j 30 at its input. Match it with a 50 line using short-circuited shunt stubs. Determine (a) the required stub admittance, (b) the distance between the stub and the antenna, (c) the stub length, and (d) the VSWR on each section of the circuit. 6.6. A lossless 100- transmission line is to be matched with a 100 +j 100 load using a double-stub tuner (Figure P6.6). Separation between the two stubs is λ/8 and the characteristic impedance is 100 . A load is connected right at the location of the first stub. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs.

228

IMPEDANCE-MATCHING NETWORKS

100 + j100 Ω 100 Ω

λ/8

Figure P6.6

6.7. A lossless 75- transmission line is to be matched with a 150 +j 15- load using a shunt-connected double-stub tuner. Separation between the two stubs is λ/8 and the characteristic impedance is 75 . The stub closest to the load (first stub) is λ/2 away from it. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs. 6.8. A lossless 50- transmission line is to be matched with a 25 +j 50- load using a double-stub tuner. Separation between the two stubs is 3λ/8 and the characteristic impedance is 50 . A load is connected at 0.2λ from the first stub, as shown in Figure P6.8. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs.

0.2λ

3λ/8

Z0 = 50 Ω Z0 = 50 Ω

Z0 = 50 Ω

ZL = 25 + j50 Ω Z0 = 50 Ω

l2

l1

Figure P6.8

229

PROBLEMS

6.9. Complete or verify the two interstage designs of 1 GHz shown as (a) and (b) in Figure P6.9.

Γin = 0.29 ∠40 °

Matching network

ΓL = 0.58 ∠20 ° O.C.

~0.174λ

~0.086λ

(a)

(b)

Figure P6.9

6.10. Determine the shortest length d in the two networks in Figure P6.10 to match the load to a 50- lossless line. Also find the stub length l for circuit (a) and the required inductance L for circuit (b).

d

O.C. stub

d

ZL

ZL

L

ΓL = 0.3 ∠45° (a)

ΓL = 0.3 ∠45° (b)

Figure P6.10

230

IMPEDANCE-MATCHING NETWORKS

6.11. Determine the shortest length d in the two networks in Figure P6.11 to match a 100 −j 50- load to a 50- lossless line. Also find the required capacitance C for circuit (a) and the stub length l for circuit (b). d

d

50 Ω

C

ZL

50 Ω

ZL

l (a)

(b)

Figure P6.11

6.12. Complete or verify the two interstage designs of 1 GHz shown as (a) and (b) in Figure P6.12. Assume that the characteristic impedance is 50 .

Matching network

Γin = 1/3 ∠−120°

ΓL = 2/3 ∠50° O.C. C

Series stub

d

~ 0.1λ (b)

(a)

Figure P6.12

6.13. Complete or verify the interstage designs at f = 4 GHz, shown as (a) and (b) in Figure P6.13.

231

PROBLEMS

Matching network

Γin = 0.3 ∠45°

Load

ΓL = 0.3 ∠−90° d

d

Γin = 0.3 ∠45° L

Γin = 0.3 ∠45° C

ZL

ZL

(b)

(a)

Figure P6.13

6.14. Complete or verify the interstage designs at f = 4 GHz shown as (a) and (b) in Figure P6.14. The characteristic impedance is 50 .

Γin = 3/7 ∠80°

ΓL = 5/9∠−150° 0.126λ

~0.385λ

C Γin = 3/7 ∠80°

L ZL

Γin = 3/7 ∠ 80°

(a)

ZL

(b)

Figure P6.14

6.15. Two types of L-section matching networks are shown in Figure P6.15. Select one that can match the load ZL = 20 −j 100 to a 50- transmission line. Find the element values at 500 MHz.

232

IMPEDANCE-MATCHING NETWORKS C Ls

Lp

50 Ω

ZL

50 Ω

L

(a)

ZL

(b)

Figure P6.15

6.16. Two types of L-section matching networks are shown in Figure P6.16. Select one that can match a 30 +j 50- load to a 50- line at 1 GHz.

C1 C 50 Ω

L

ZL

ZL

50 Ω C2

(a)

(b)

Figure P6.16

6.17. Design a matching network (Figure P6.17) that will transform a 50 −j 50- load to the input impedance of 25 +j 25 .

Zin = 25 + j25 Ω

Matching network

ZL = 50 − j50 Ω

Figure P6.17

6.18. Match the load in Figure P6.18 to a 50- generator using lumped elements. Assume that the signal frequency as 4 GHz.

Γin = 0

Matching network

Load

ΓL = 0.5 ∠45°

Figure P6.18

233

PROBLEMS

6.19. Two types of L-section matching networks are shown in Figure P6.19. Select one that can match the load ZL = 60 −j 20 to a 50- transmission line. Find the element values at 500 MHz. Ls

50 Ω

Ls

Cp

ZL

50 Ω

(a)

Cp

ZL

(b)

Figure P6.19

6.20. Design a lumped-element network that will provide a load ZL = 30 +j 50 for the amplifier shown in Figure P6.20 operating at 2 GHz.

Lumped elements

Amplifier

50 Ω

ZL

Figure P6.20

6.21. Design a lumped-element network that will provide a load ZL = 30 +j 50 for the amplifier shown in Figure P6.21, and stop the biasing voltage from appearing across 50 . The amplifier is operating at 500 MHz.

Lumped elements

Amplifier

50 Ω

ZL

Figure P6.21

6.22. A 200- load is to be matched with a 50- line. Design (a) a resistive network and (b) a reactive network to match the load. (c) Compare the performance of the two designs.

7 IMPEDANCE TRANSFORMERS

In Chapter 6, several techniques were considered to match a given load impedance at a fixed frequency. These techniques included transmission line stubs as well as lumped elements. Note that lumped-element circuits may not be practical at higher frequencies. Further, it may be necessary in certain cases to keep the reflection coefficient below a specified value over a given frequency band. In this chapter we present transmission line impedance transformers that can meet such requirements. The chapter begins with a single-section impedance transformer that provides perfect matching at a single frequency. A matching bandwidth can be increased at the cost of a higher reflection coefficient. This concept is used to design multisection transformers. The characteristic impedance of each section is controlled to obtain the desired passband response. Multisection binomial transformers exhibit an almost flat reflection coefficient about the center frequency and increase gradually on either side. A wider bandwidth is achieved with an increased number of quarter-wave sections. Chebyshev transformers can provide even wider bandwidth with the same number of sections, but the reflection coefficient exhibits ripples in its passband. The chapter includes a procedure to design these multisection transformers as well as transmission line tapers. The chapter concludes with a brief discussion on the Bode–Fano constraints, which provide an insight into the trade-off between the bandwidth and the allowed reflection coefficient. 7.1 SINGLE-SECTION QUARTER-WAVE TRANSFORMERS We considered a single-section quarter-wavelength transformer design problem in Example 3.5. In this section we present a detailed analysis of such circuits. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

234

SINGLE-SECTION QUARTER-WAVE TRANSFORMERS

235

l

Z0 Z1

Zin

Figure 7.1

RL

Single-section quarter-wave transformer.

Consider the load resistance RL that is to be matched with a transmission line of characteristic impedance Z0 . Assume that a transmission line of length l and characteristic impedance Z1 is connected between the two, as shown in Figure 7.1. Its input impedance Zin is found as follows: Zin = Z1

RL + j Z1 tan βl Z1 + j RL tan βl

(7.1.1)

√ For βl = 90◦ (i.e., l = λ/4) and Z1 = Z0 RL , Zin is equal to Z0 ; hence, there is no reflected wave beyond this point toward the generator. However, it reappears at other frequencies when βl = 90◦ . The corresponding reflection coefficient in can be determined as follows: in = =

Zin − Z0 Z1 [(RL + j Z1 tan βl)/(Z1 + j RL tan βl)] − Z0 = Zin + Z0 Z1 [(RL + j Z1 tan βl)/(Z1 + j RL tan βl)] + Z0 RL − Z0 = ρin exp(j ϕ) √ RL + Z0 + j 2 Z0 RL tan βl

Therefore, ρin = =

[(RL + Z0

RL − Z0 + 4Z0 RL tan2 βl]1/2

)2

1 √ {1 + [2 Z0 RL /(RL − Z0 ) sec βl]2 }1/2

(7.1.2)

Variation in ρin with frequency is illustrated in Figure 7.2. For βl near 90◦ , it can be approximated as follows: |RL − Z0 | |RL − Z0 | ρin ≈ √ cos βl ≈ √ 2 Z0 RL tan βl 2 Z0 RL

(7.1.3)

236

IMPEDANCE TRANSFORMERS 0.35 0.30 0.25

ρin

0.20 0.15 0.10 0.05 0.00 0.0

ρM

0.5

θ1 1.0

π − θ1 1.5

2.0

2.5

3.0

3.5

βl

Figure 7.2 Reflection coefficient characteristics of a single-section impedance transformer used to match a 100- load to a 50- line.

If ρM is the maximum allowable reflection coefficient at the input, then √ 2ρM Z0 RL π θ1 < (7.1.4) cos θ1 = 2 (RL − Z0 ) 1 − ρ2M In the case of a TEM wave propagating on the transmission line, βl = (π/2) (f/f0 ), where f0 is the frequency at which βl = π/2. In this case, the bandwidth f2 − f1 = f is given by 2f0 (7.1.5) f = (f2 − f1 ) = 2(f0 − f1 ) = 2 f0 − θ1 π and the fractional bandwidth is

√ 2ρM Z0 RL f 4 −1 = 2 − cos f0 π 2 (RL − Z0 ) 1 − ρM

(7.1.6)

Example 7.1 Design a single-section quarter-wave impedance transformer to match a 100- load to a 50- air-filled coaxial line at 900 MHz (Figure 7.3). Determine the range of frequencies over which the reflection coefficient remains below 0.05.

237

MULTISECTION QUARTER-WAVE TRANSFORMERS

Z0

Z1

RL

Γin

Figure 7.3

Setup for Example 7.1.

SOLUTION For RL = 100 and Z0 = 50 , Z1 = and l=

√

100 × 50 = 70.7106781

λ 3 × 108 m = 8.33 cm = 4 4 × 900 × 106

The magnitude of the reflection coefficient increases as βl changes from π/2 (i.e., the signal frequency changes from 900 MHz). If the maximum allowed ρ is ρM = 0.05 (VSWR = 1.1053), the fractional bandwidth is found to be √ 2ρM Z0 RL 4 f −1 = 2 − cos = 0.180897 f0 π 2 (RL − Z0 ) 1 − ρM Therefore, 818.5964 MHz ≤ f ≤ 981.4037 MHz or 1.4287 ≤ βl ≤ 1.7129.

7.2 MULTISECTION QUARTER-WAVE TRANSFORMERS Consider an N -section impedance transformer connected between a transmission line of characteristic impedance of Z0 and load RL , as shown in Figure 7.4. As indicated, the length of every section is the same, although their characteristic impedances are different. Impedance at the input of N th section can be found as follows: exp(j βl) + N exp(−j βl) N Zin = ZN (7.2.1) exp(j βl) − N exp(−j βl) where N =

RL − ZN RL + ZN

(7.2.2)

The reflection coefficient seen by the (N − 1)st section is N−1 =

N Zin ZN (ej βl + N e−j βl ) − ZN−1 (ej βl − N e−j βl ) − ZN−1 = N ZN (ej βl + N e−j βl ) + ZN−1 (ej βl − N e−j βl ) Zin + ZN−1

238

IMPEDANCE TRANSFORMERS

x l l l ZN−1

ZN

RL

Z2 Z1 Z0

Figure 7.4 N-section impedance transformer.

or N−1 =

(ZN − ZN−1 )ej βl + N (ZN + ZN−1 )e−j βl (ZN + ZN−1 )ej βl + N (ZN − ZN−1 )e−j βl

Therefore, N−1 =

N−1 + N e−j 2βl 1 + N N−1 e−j 2βl

(7.2.3)

ZN − ZN−1 ZN + ZN−1

(7.2.4)

where N−1 =

If ZN is close to RL and ZN−1 is close to ZN , then N and N−1 are small quantities, and a first-order approximation can be assumed. Hence, N−1 ≈ N−1 + N e−j 2βl

(7.2.5)

Similarly, N−2 ≈ N−2 + N e−j 2βl = N−2 + N−1 e−j 2βl + N e−j 4βl

Therefore, by induction, the reflection coefficient seen by the feeding line is ≈ 0 + 1 e−j 2βl + 2 e−j 4βl + · · · + N−1 e−j 2(N−1)βl + N e−j 2Nβl or =

N

(7.2.6)

n e−j 2nβl

(7.2.7)

Zn+1 − Zn Zn+1 + Zn

(7.2.8)

n=0

where n =

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

239

Thus, we need a procedure to select n so that is minimized over the desired frequency range. To this end, we recast equation (7.2.6) as follows: = 0 + 1 w + 2 w2 + · · · + N wN = N

N

(w − wn )

(7.2.9)

n=1

where ϕ = −2βl

(7.2.10)

w = ej ϕ

(7.2.11)

and

Note that for βl = 0 (i.e., λ → ∞), individual transformer sections in effect have no electrical length and the load RL appears to be connected directly to the main line. Therefore, N RL − Z0 = n = (∵ w = 1) (7.2.12) RL + Z0 n=0 and only N of the (N + 1)-section reflection coefficients can be selected independently.

7.3 TRANSFORMER WITH UNIFORMLY DISTRIBUTED SECTION REFLECTION COEFFICIENTS If all of the section reflection coefficients are equal, (7.2.9) can be simplified as follows: wN+1 − 1 = 1 + w + w2 + w3 + · · · + wN−1 + wN = N w−1

(7.3.1)

or ej (N+1)ϕ − 1 sin{[(N + 1)/2]ϕ} = = ej Nϕ/2 j ϕ N e −1 sin(ϕ/2) Hence, sin{[(N + 1)/2]ϕ} = (N + 1)ρN sin{[(N + 1)/2]ϕ} || = ρ(ϕ) = ρN sin(ϕ/2) (N + 1) sin(ϕ/2) (7.3.2) and from (7.2.12), N RL − Z0 n = (N + 1)ρN = (7.3.3) RL + Z0 n=0

240

IMPEDANCE TRANSFORMERS

Therefore, equation (7.3.2) can be written as follows: RL − Z0 sin[(N + 1)βl] · ρ(βl) = RL + Z0 (N + 1) sin βl

(7.3.4)

This can be viewed as an equation that describes magnitude ρ of the reflection coefficient as a function of frequency. As (7.3.4) indicates, a pattern of ρ(βl) repeats periodically with an interval of π. It peaks at nπ, where n is an integer including zero. Further, there are N − 1 minor lobes between two consecutive main peaks. The number of zeros between the two main peaks of ρ(βl) is equal to the number of quarter-wave sections, N . Consider that there are three quarter-wave sections connected between a 100- load and a 50- line. Its reflection coefficient characteristics can be found from (7.3.4), as illustrated in Figure 7.5. There are three zeros in it, one at βl = π/2 and the other two located symmetrically around this point. In other words, zeros occur at βl = π/4, π/2, and 3π/4. When the number of quarterwave sections is increased from three to six, the ρ(βl) plot changes as illustrated in Figure 7.6. For a six-section transformer, Figure 7.6 shows five minor lobes between two main peaks of ρ(βl). One of these minor lobes has its maximum value (peak) at βl = π/2. Six zeros of this plot are located symmetrically: βl = nπ/7, n = 1, 2, . . . , 6. Thus, characteristics of ρ(βl) can be summarized as follows: ž ž

The pattern of ρ(βl) repeats with an interval of π. There are N nulls and N − 1 minor peaks in an interval. 0.35 0.30 0.25

ρ(βl)

0.20 0.15 0.10 0.05 0.00 0.0

P1 0.5

P2 1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.5 Reflection coefficient versus βl of a three-section transformer with equal section reflection coefficients for RL = 100 and Z0 = 50 .

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

241

0.35 0.30 0.25

ρ(βl)

0.20 0.15 0.10 0.05

P1

P2

0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.6 Reflection coefficient versus βl for a six-section transformer with equal section-reflection coefficients (RL = 100 and Z0 = 50 ). ž ž

When N is odd, one of the nulls occurs at βl = π/2 (i.e., l = λ/4). If ρM is specified as an upper bound on ρ to define the frequency band, points P1 and P2 bound the acceptable range of βl. This range becomes larger as N increases.

Since wN+1 − 1 =

N

(w − wn )

(7.3.5)

n = 1, 2, . . . , N

(7.3.6)

n=0

where wn = ej [2πn/(N+1)] (7.3.1) may be written as follows: N N = (w − wn ) = (w − ej [2πn/(N+1)] ) N n=1 n=1

(7.3.7)

This equation is of the form of (7.2.9). It indicates that when section reflection coefficients are the same, roots are equispaced around the unit circle on the complex w-plane with the root at w = 1 deleted. This is illustrated in Figure 7.7 for N = 3. It follows that N N ρ j [2πn/(N+1)]) = (w − e |(w − ej [2πn/(N+1)] )| (7.3.8) = ρN n=1

n=1

242

IMPEDANCE TRANSFORMERS Imaginary w

1 ϕ Real w

Figure 7.7

Location of zeros on a unit circle on the complex w-plane for N = 3. w3

d3

w2

d2

d1

w1

Figure 7.8

Graphical representation of (7.3.9) for N = 3 and ϕ = 0.

Since w = ej ϕ is constrained to lie on the unit circle, the distance between w and wn is given by dn (ϕ) = |ej ϕ − ej [2πn/(N+1)] | (7.3.9) For the case of N = 3, dn (ϕ = 0) is illustrated in Figure 7.8. From (7.3.3), (7.3.8), and (7.3.9), we get ρ(ϕ) =

N 1 RL − Z0 dn (ϕ) N + 1 RL + Z0 n=1

(7.3.10)

Thus, as ϕ = −2βl varies from 0 to 2π, w = ej ϕ makes one complete traverse of the unit circle, and distances d1 , d2 , . . . , dN vary with ϕ. If w coincides with the

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

243

root wn , the distance dn is zero. Consequently, the product of the distances is zero. Since the product of these distances is proportional to the reflection coefficient, ρ(ϕn ) goes to zero. It attains a local maximum whenever w is approximately halfway between successive roots. Example 7.2 Design a four-section quarter-wavelength impedance transformer with uniform distribution of reflection coefficient to match a 100- load to a 50- air-filled coaxial line at 900 MHz. Determine the range of frequencies over which the reflection coefficient remains below 0.1. Compare this bandwidth with that obtained for a single-section impedance transformer. SOLUTION From (7.3.3) with N = 4, we have 4 = and from (7.2.2),

1 1 100 − 50 = 5 100 + 50 15

1 100 − Z4 = 15 100 + Z4

or 100 + Z4 = 1500 − 15Z4 ⇒ 16Z4 = 1400 ⇒ Z4 = 87.5 The characteristic impedance of other sections can be determined from (7.2.8) as follows: 1 Z4 − Z3 = ⇒ 87.5 + Z3 = 15 × 87.5 − 15Z3 ⇒ Z3 = 76.5625 15 Z4 + Z3 Z3 − Z2 14 × 76.5625 1 = = 66.9922 ≈ 67 ⇒ Z2 = 15 Z3 + Z2 16 and Z2 − Z1 14 × 67 1 = = 58.625 ⇒ Z1 = 15 Z2 + Z1 16 The frequency range over which reflection coefficient remains below 0.1 is determined from (7.3.4) as follows: 1 sin 5θm sin 5θm = 1.5 0.1 = × ⇒ 3 5 sin θm sin θm where θm represents the value of βl at which magnitude of reflection coefficient is equal to 0.1. This is a transcendental equation that can be solved graphically. To that end, one needs to plot | sin 5θm | and 1.5 | sin θm | versus θm on the same graph and look for the intersection of two curves. Alternatively, a numerical

244

IMPEDANCE TRANSFORMERS 0.35 0.30 N=1

0.25 N=4 0.20

θ1

0.10

θ4

ρ

0.15

0.05 0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.9 Reflection coefficient versus βl for a four-section (N = 4) and a single section (N = 1) of impedance transformer.

method can be employed to determine θm . Two solutions to this equation are found to be 0.476 and 2.665, respectively. Hence, 0.476 ≤ βl ≤ 2.665 or

272.7 MHz ≤ f ≤ 1.5269 GHz

Corresponding bandwidth with a single section can be evaluated as follows: 1.2841 ≤ βl ≤ 1.8575 or

735.74 MHz ≤ f ≤ 1.0643 GHz

Reflection coefficient characteristics for two cases are displayed in Figure 7.9. Clearly, it has a much wider bandwidth, with four sections in comparison with that of a single section.

7.4 BINOMIAL TRANSFORMERS As shown in Figures 7.5 and 7.6, there are peaks and nulls in the passband of a multisection quarter-wavelength transformer with uniform section reflection coefficients. This characteristic can be traced to equispaced roots on the unit circle. One way to avoid this behavior is to place all the roots at a common pointw equal to −1. With this setting, distances dn are the same for all cases and N n=1 (w − wn ) goes to zero only once. It occurs for ϕ equal to −π (i.e., at βl = π/2). Thus, ρ is zero only for the frequency at which each section of

245

BINOMIAL TRANSFORMERS

the transformer is λ/4 long. With wn = −1 for all n, equation (7.2.9) may be written as N! wm = (w − wn ) = (w + 1)N = N m!(N − m)! n=1 m=0 N

N

(7.4.1)

The following binomial expansion is used in writing (7.4.1): (1 + x) = m

N n=1

m! xn n!(m − n)!

A comparison of equation (7.4.1) with (7.2.7) indicates that n N! = N n!(N − n)!

n = 0, 1, 2, . . . , N

(7.4.2)

and therefore the section reflection coefficients, normalized to N , are binomially distributed. From equation (7.4.1), = N (w + 1)N ⇒ (βl) = N (e−j 2βl + 1)N = N 2N e−j Nβl (cos βl)N or ρ(βl) = ρN 2N | cos βl|N

(7.4.3)

For βl = 0, load RL is effectively connected to the input line. Therefore, RL − Z0 ρ(0) = ρN 2 = RL + Z0 N

and

RL − Z0 × | cos βl|N ρ(βl) = RL + Z0

(7.4.4)

(7.4.5)

Reflection coefficient characteristics of multisection binomial transformers versus βl (in degrees) are illustrated in Figure 7.10. The reflection coefficient scale is normalized with the load reflection coefficient ρL . Unlike the preceding case of uniformly distributed section reflection coefficients, it shows a smooth characteristic without lobes. Example 7.3 Design a four-section quarter-wavelength binomial impedance transformer to match a 100- load to a 50- air-filled coaxial line at 900 MHz. Determine the range of frequencies over which the reflection coefficient remains below 0.1. Compare this result with that obtained in Example 7.2.

246

IMPEDANCE TRANSFORMERS 1.0 N=1 N = 20

Reflection coefficient

0.8 N=2

N=9

0.6 N=6 0.4

0.2

0.0

0

Figure 7.10

45

90 θ (deg)

135

180

Reflection coefficient versus βl for a multisection binomial transformer.

SOLUTION With N = 4 and n = 0, 1, and 2 in (7.4.2), we get 4! 0 4 = ⇒ 0 = 4 = 4 0!4! 4 3 4! 1 = = ⇒ 1 = 3 = 44 4 1!3! 4 and 4! 2 = = 6 ⇒ 2 = 64 4 2!2! From (7.4.4), RL − Z0 ⇒ ρ4 = 1 ρ(0) = ρN 2 = RL + Z0 24 N

100 − 50 1 100 + 50 = 48 = 0.020833

and from (7.2.8), n =

Zn+1 − Zn ⇒ ρn (Zn+1 + Zn ) = Zn+1 − Zn Zn+1 + Zn

Therefore, Zn =

1 − ρn × Zn+1 1 + ρn

(7.4.6)

1 + ρn × Zn 1 − ρn

(7.4.7)

Alternatively, Zn+1 =

BINOMIAL TRANSFORMERS

247

Characteristic impedance of each section can be determined from (7.4.6), as follows: Z4 =

1 − 1/48 × 100 = 95.9184 1 + 1/48

Z3 =

1 − 4/48 × 95.9184 = 81.1617 1 + 4/48

Z2 =

1 − 6/48 × 81.1617 = 63.1258 1 + 6/48

Z1 =

1 − 4/48 × 63.1258 = 53.4141 1 + 4/48

and

If we continue with this formula, we find that Z0 =

1 − 1/48 × 53.4141 = 51.2339 1 + 1/48

This is different from the given value of 50 . It happened because of the approximation involved in the formula. Error keeps building up if the characteristic impedances are determined proceeding in just one way. To minimize it, common practice is to determine the characteristic impedances up to halfway proceeding from the load side and then the remaining half from the input side. Thus, Z1 and Z2 should be determined from (7.4.7) as follows: Z1 =

1 + 1/48 × 50 = 52.1277 1 − 1/48

Z2 =

1 + 4/48 × 50 = 61.6054 1 − 4/48

and

The frequency range over which the reflection coefficient remains below 0.1 can be determined from (7.4.5) as follows: 0.1 = 13 | cos(ϑM )|4 ⇒ ϑM = 0.7376 Therefore, 0.7376 ≤ βl ≤ 2.404

or 422.61 MHz ≤ f ≤ 1.3774 GHz

Clearly, it has a larger frequency bandwidth than that of a single section. However, it is less than the bandwidth obtained with uniformly distributed section reflection coefficients. Reflection coefficient as a function of βl is illustrated in Figure 7.11.

248

IMPEDANCE TRANSFORMERS 0.35 0.30 0.25

ρ(βl)

0.20 θB

0.15 0.10 0.05 0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.11

Reflection coefficient of a four-section binomial transformer versus βl.

7.5 CHEBYSHEV TRANSFORMERS Consider again the case of a uniform three-section impedance transformer that is connected between a 50- line and 100- load. Distribution of zeros around the unit circle is shown in Figure 7.12 with solid points. Its frequency response is illustrated in Figure 7.13 as curve (a). Now, let us move two of these zeros to ±120◦ while keeping the remaining one fixed at 180◦ , as illustrated by unfilled Imaginary w

Real w

Figure 7.12 Distribution of zeros for a uniform three-section impedance transformer (solid), with two of those zeros moved to ±120◦ (unfilled) or to ±60◦ (hatched).

249

CHEBYSHEV TRANSFORMERS 0.5

0.4 (a) 0.3

(b)

ρ(ϕ)

(c) 0.2

0.1

0 0

1

2

3

4

5

6

ϕ

Figure 7.13 Reflection coefficient versus ϕ for (a) a uniform three-section transformer, (b) two zeros moved to ±120◦ , and (c) two zeros moved to ±60◦ .

points in Figure 7.12. With this change in the distribution of zeros on the complex w-plane, ρ(βl) versus βl varies as shown by curve (b) in Figure 7.13. It shows relatively much lower peaks of intervening lobes, while the widths and heights of the two main lobes increase. On the other hand, if we move the two zeros to ±60◦ , the main peaks go down while intervening lobes rise. This is illustrated by hatched points in Figure 7.12 and by curve (c) in Figure 7.13. In this case, minor lobes increase while the main lobes are reduced in size. Note that zeros in this case are uniformly distributed around the unit circle, and therefore its ρ(βl) characteristic has identical lobes. Thus, the heights of intervening lobes decrease at the cost of the main lobe when zeros are moved closer together. On the other hand, moving the zeros farther apart raises the level of intervening lobes but reduces the main lobe. However, we need a systematic method to determine the location of each zero for a maximum permissible reflection coefficient, ρM , and the number of quarter-wave sections, N . An optimal distribution of zeros around the unit circle will keep the peaks of all passband lobes at the same height of ρM . In order to have the magnitudes of all minor lobes in the passband equal, section reflection coefficients are determined by the characteristics of Chebyshev functions, named after the Russian mathematician who first studied them. These functions satisfy the following differential equation: (1 − x 2 )

d 2 Tm (x) dTm (x) −x + m2 Tm (x) = 0 2 dx dx

(7.5.1)

Chebyshev functions of degree m, represented by Tm (x), are mth-degree polynomials that satisfy (7.5.1). The first four of these and a recurrence relation for

250

IMPEDANCE TRANSFORMERS

higher-order Chebyshev polynomials are given as follows: T1 (x) = x T2 (x) = 2x 2 − 1 T3 (x) = 4x 3 − 3x T4 (x) = 8x 4 − 8x 2 + 1 .. . Tm (x) = 2xTm−1 (x) − Tm−2 (x) Alternatively, −1 cos(m cos x) − 1 ≤ x ≤ 1 Tm (x) = cosh(m cosh−1 x) x ≥ 1 (−1)m cosh(m cosh−1 |x|) x ≤ −1 Note that for x = cos θ,

(7.5.2)

Tm (cos θ) = cos mθ

(7.5.3)

Figure 7.14 depicts Chebyshev polynomials of degrees 1 through 4. The following characteristics can be noted: ž ž

The magnitudes of these polynomials oscillate between ±1 for −1 ≤ x ≤ 1. For |x| > 1, |Tm (x)| increases at a faster rate with x as m increases.

6.0 5.0 4.0 3.0 m=4

2.0

m=3

m=2

m=1

1.0 0.0 −1.0 −2.0 −3.0 −1.5

−x0 −1.0

x0 −0.5

0.0

0.5

1.0

Figure 7.14 Chebyshev polynomials for m = 1, 2, 3, 4.

1.5

251

CHEBYSHEV TRANSFORMERS ž

ž

The numbers of zeros are equal to the order of the polynomials. Zeros of an even-order polynomial are located symmetrically about the origin, with one of the minor lobes’ peak at x = 0. Polynomials of odd orders have one zero at x = 0 while the remaining zeros are located symmetrically.

These characteristics of Chebyshev polynomials are utilized to design an impedance transformer that has ripples of equal magnitude in its passband. The number of quarter-wave sections determines the order of Chebyshev polynomials and the distribution of zeros on the complex w-plane is selected according to that. With x0 properly selected, |Tm (x)| corresponds precisely to ρ(βl). This is done by linking βl to x of Chebyshev polynomial as follows: x = x0 cos βl

(7.5.4)

Consider the design of a three-section equal-ripple impedance transformer. A Chebyshev polynomial of order 3 is appropriate for this case. Figure 7.15 illustrates T3 (x) together with (7.5.4). It is to be noted that Chebyshev variable x and angle ϕ on the complex w-plane are related through (7.5.4) because ϕ is equal to −2βl. As ϕ varies from 0 to −2π, x changes as illustrated in Figure 7.15(b) and the corresponding T3 (x) in Figure 7.15(a). Figure 7.16 shows |T3 (ϕ)|, which can represent the desired ρ(ϕ) provided that (RL − Z0 ) (R + Z ) T3 (ϕ = 0) T3 (x0 ) L 0 = = 1 1 ρM

(7.5.5)

where ρM is the maximum allowed reflection coefficient in the passband. For an m-section impedance transformer, (7.5.5) can be written as RL − Z0 1 Tm (x0 ) = RL + Z0 ρM

(7.5.6)

Location x0 can now be determined from (7.5.2) as follows:

1 −1 x0 = cosh × cosh Tm (x0 ) m

(7.5.7)

With x0 determined from (7.5.7), |Tm (x)| represents the ρ(βl) desired. Hence, zeros of the reflection coefficient are the same as those of Tm (x). Since Chebyshev polynomials have zeros in the range −1 < x < 1, zeros of ρ(ϕ) can be determined from (7.5.2). Therefore,

π Tm (xn ) = 0 ⇒ cos(m × cos−1 xn ) = 0 = ± cos (2n − 1) 2

n = 1, 2, . . .

252

IMPEDANCE TRANSFORMERS 2.0

(a)

0.0

−2.0 (b)

x0 cos βl

0

βl π

−1.5

−1.0

−0.5

0.0

0.5

1.0

1.5

Figure 7.15 Third-order Chebyshev polynomial (a) and its variable, x versus βl (b). 2.5

T3(x0)

RL − Z0 2.0 RL + Z 0 1.5

ρM 1.0

0.5

0.0 −7.0

Figure 7.16

−6.0

−5.0

−4.0

−3.0

−2.0

−1.0

0.0

|T3 (x)| versus ϕ and its relationship with the desired ρ(ϕ).

where xn is the location of the nth zero. Hence,

π xn = ± cos (2n − 1) 2m

(7.5.8)

and the corresponding ϕn can be evaluated from (7.5.4) as follows: ϕn = 2 cos−1

xn x0

(7.5.9)

253

CHEBYSHEV TRANSFORMERS

The zeros of ρ(ϕ), wn , on the complex w-plane are now known because wn = eiϕn . Equation (7.2.9) can be used to determine the section reflection coefficient, n . Zn is, in turn, determined from equation (7.2.8). The bandwidth of the impedance transformer extends from x = −1 to x = +1. With βl = θM at x = 1, (7.5.4) gives 1 cos θM = x0 Therefore, bandwidth of a Chebyshev transformer can be expressed as follows: cos−1

1 1 ≤ βl ≤ π − cos−1 x0 x0

(7.5.10)

Example 7.4 Reconsider the matching of a 100- load to a 50- line. Design an equal-ripple four-section quarter-wavelength impedance transformer and compare its bandwidth with those obtained earlier for ρM = 0.1. SOLUTION From (7.5.6), 100 − 50 × 1 = 1 = 3.3333 T4 (x0 ) = 100 + 50 0.1 0.3 Therefore, x0 can now be determined from (7.5.7) as follows: x0 = cosh 14 cosh−1 (3.3333) In case inverse hyperbolic functions are not available on a calculator, the following procedure can be used to evaluate x0 . Assume that y = cosh−1 (3.3333) ⇒ 3.3333 = cosh y = (ey + e−y )/2. Hence, ey + e−y = 2 × 3.3333 = 6.6666 or e

2y

− 6.6666 × e + 1 = 0 ⇒ e = y

y

6.6666 ±

(6.6666)2 − 4 = 6.5131 2

and 0.1535y = ln(6.5131) = 1.8738 Therefore,

1.8738 1 cosh−1 (3.3333) = = 0.4684 4 4

and x0 = cosh(0.4684) =

e0.4684 + e−0.4684 = 1.1117 2

Note that the other solution of ey produces y = −1.874 and x0 = 1.1117.

254

IMPEDANCE TRANSFORMERS

From (7.5.8) and (7.5.9), we get xn = ±0.9239, ±0.3827 and

◦

◦

◦

◦

ϕn = −67.61 , −139.71 , −220.29 , −292.39 Therefore, w1 = 0.381 − j 0.925 w2 = −0.763 − j 0.647 w3 = −0.763 + j 0.647 and w4 = 0.381 + j 0.925 Now, from equation (7.2.9) we get

= 4 (1 + 0.764w + 0.837w2 + 0.764w3 + w4 ) Therefore, 0 = 4 , 1 = 3 = 0.7644

and 2 = 0.8374

From equation (7.2.12), we have =

N

n =

RL − Z0 RL − Z0 ⇒ 4.3654 = RL + Z0 RL + Z0

4 =

100 − 50 1 × = 0.076 4.365 100 + 50

n=0

Hence, 0 = 4 = 0.076,

1 = 3 = 0.058,

and 2 = 0.064

Now, the characteristic impedance of each section can be determined from (7.2.8) as follows: ZL − Z4 1 − 4 ⇒ Z4 = ZL = 85.87 ZL + Z4 1 + 4 Z4 − Z3 1 − 3 ⇒ Z3 = Z4 = 76.46 3 = Z4 + Z3 1 + 3 Z3 − Z2 1 − 2 2 = ⇒ Z2 = Z3 = 67.26 Z3 + Z2 1 + 2

4 =

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

255

and 1 =

Z2 − Z1 1 − 1 ⇒ Z1 = Z2 = 59.89 Z2 + Z1 1 + 1

To minimize the accumulating error in Zn , it is advisable to calculate half of the impedance values from the load side and the other half from the input side. In other words, Z1 and Z2 should be determined as follows: Z1 =

1 + 0 Z0 = 58.2251 1 − 0

Z2 =

1 + 1 Z1 = 65.3951 1 − 1

and

The bandwidth is determined from (7.5.10) as follows: cos βl =

1 = 0.899 ⇒ βl = 0.452 x0

Therefore, 0.452 ≤ βl ≤ 2.6896 Thus, the bandwidth is greater than what was achieved from either a uniform or a binomial distribution of n coefficients.

7.6 EXACT FORMULATION AND DESIGN OF MULTISECTION IMPEDANCE TRANSFORMERS The analysis and design presented so far is based on the assumption that the section reflection coefficients are small, as implied by (7.2.5). If this is not the case, an exact expression for the reflection coefficient must be used. Alternatively, there are design tables available in the literature (e.g., Matthaei et al., 1980) that can be used to synthesize an impedance transformer. In case of a two- or threesection Chebyshev transformer, the design procedure summarized below may be used as well. Exact formulation of the multisection impedance transformer is conveniently developed via the power loss ratio, PLR , defined as follows: PLR =

incident power power delivered to the load

If Pin represents the incident power and ρin is the input reflection coefficient, then Pin 1 PLR − 1 = ⇒ ρin = (7.6.1) PLR = PLR (1 − ρ2in )Pin 1 − ρ2in

256

IMPEDANCE TRANSFORMERS

For any transformer, ρin can be determined from its input impedance Zin . The power loss ratio PLR can subsequently be evaluated from (7.6.1). This PLR can be expressed in terms of Q2N (cos θ), an even polynomial of degree 2N in cos θ. Hence, PLR = 1 + Q2N (cos θ) (7.6.2) Coefficients of Q2N (cos θ) are functions of various impedances Zn . For an equalripple characteristic in the passband, a Chebyshev polynomial can be used to specify PLR as follows: PLR = 1 + k 2 TN2 (sec θM cos θ)

(7.6.3)

where k 2 is determined from the maximum value of PLR in the passband. θM represents the value of βl that corresponds to the maximum allowed reflection coefficient ρM . Since TN2 (sec θM cos θ) has a maximum value of unity in the passband, the extreme value of PLR will be 1 + k 2 . Various characteristic impedances are determined by equating (7.6.2) and (7.6.3). Further, (7.6.1) produces a relation between ρM and k 2 , as follows: k2 ρM = (7.6.4) 1 + k2 For the two-section impedance transformer shown in Figure 7.17, reflection coefficient characteristics as a function of θ are illustrated in Figure 7.18. The power loss ratio for this transformer is found to be PLR = 1 +

(ZL − Z0 )2 (sec2 θz cos2 θ − 1)2 4ZL Z0 tan4 θz

(7.6.5)

where θz is the value of θ at the lower zero (i.e., θz < π/2), as shown in Figure 7.18. The maximum power loss ratio, PLR max , is found to be PLR

max

Z0

=1+

Z1

(ZL − Z0 )2 cot4 θz 4ZL Z0

Z2

(7.6.6)

ZL

θ θ

Figure 7.17

Two-section impedance transformer.

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

257

|Γin |

ρM

θ θM

θZ

π/2

π − θM

Figure 7.18 Reflection coefficient characteristics of a two-section Chebyshev transformer.

Hence, k2 =

(ZL − Z0 )2 cot4 θz 4ZL Z0

(7.6.7)

Required values of characteristic impedances, Z1 and Z2 , are determined from the equations Z12 = Z02

ZL (ZL − Z0 )2 + 2 4 Z0 4Z0 tan θz

and Z2 =

1/2 +

(ZL − Z0 )Z0 2 tan2 θz

ZL Z0 Z1

(7.6.8)

(7.6.9)

The passband edge, θM , is given by √ θM = cos−1 ( 2 cos θz )

(7.6.10)

Hence, the bandwidth of this transformer is found as θM ≤ θ = βl ≤ π − θM

(7.6.11a)

258

IMPEDANCE TRANSFORMERS

For a TEM wave,

4θM f =2− f0 π

(7.6.11b)

If the bandwidth is specified in a design problem along with ZL and Z0 , then θM is known. Therefore, θz can be determined from (7.6.10). Impedances Z1 and Z2 are determined subsequently from (7.6.8) and (7.6.9), respectively. The corresponding maximum reflection coefficient ρM can easily be evaluated following k 2 from (7.6.7). On the other hand, if ρM is specified instead of θM , then k 2 is determined from (7.6.4). θz , in turn, is calculated from (7.6.7). Now, Z1 , Z2 , and θM can be determined from (7.6.8), (7.6.9), and (7.6.10), respectively. In the limit θz → π/2, two zeros of ρ in Figure 7.18 come together to give a maximally flat transformer. In that case, (7.6.8) and (7.6.9) are simplified to give Z1 = ZL × Z0

1/4

3/4

(7.6.12)

3/4 ZL

1/4 Z0

(7.6.13)

Z2 =

×

and θM = cos−1 (cot θz )

(7.6.14)

Example 7.5 Use the exact theory to design a two-section Chebyshev transformer to match a 500- load to a 100- line. The required fractional bandwidth is 0.6. What is the resulting value of ρM ? SOLUTION From (7.6.10) and (7.6.11), √ 2 − f/f0 4 1 f −1 = 2 − cos ( 2 cos θz ) ⇒ cos θz = √ cos f0 π 4/π 2 2 − 0.6 1 = 0.321 = √ cos 4/π 2 Therefore, θz = 1.244 rad and √ √ θM = cos−1 ( 2 cos θz ) = cos−1 ( 2 × 0.321) = 1.1 From (7.6.7), k2 =

(ZL − Z0 )2 (Z L − 1)2 cot4 θz = cot4 θz = 0.0106 4ZL Z0 4Z L

Therefore, from (7.6.4), ρM =

k2 = 0.1022 1 + k2

259

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

Now, from (7.6.8),

1/2 ZL (ZL − Z0 )Z0 (ZL − Z0 )2 2 + + ⇒ Z1 Z0 2 tan2 θz 4Z02 tan4 θz 1/2 (Z L − 1)2 ZL − 1 = + ZL + = 2.4776 4 tan4 θz 2 tan2 θz

Z12 = Z02

Therefore, Z1 =

√

2.4776 = 1.57 ⇒ Z1 = 157.41

and from (7.6.9), Z2 =

ZL Z0 = 317.65 Z1

Check : For θ = π/2, 2

Z in =

Z1

Z = 2 L

Z2

=

Z in − 1 Z in + 1

1.57412 (5) = 1.2277 3.17652 = 0.1022 = ρm

Let us consider the design of a three-section impedance transformer. Figure 7.19 shows a three-section impedance transformer connected between the load ZL and a transmission line of characteristic impedance Z0 . The reflection coefficient characteristic of such a Chebyshev transformer is illustrated in Figure 7.20. The power loss ratio, PLR , of a three-section Chebyshev transformer is found to be PLR = 1 +

Z0

(ZL − Z0 )2 (sec2 θz cos2 θ − 1)2 cos2 θ 4ZL Z0 tan4 θz

(7.6.15)

ZL

Z1 Z2

Z3

θ θ θ

Figure 7.19 Three-section impedance transformer.

260

IMPEDANCE TRANSFORMERS

| Γin |

ρM θ θM θZ

Figure 7.20 former.

Reflection coefficient characteristics of a three-section Chebyshev trans-

It attains the maximum allowed value at θ = θM , where 2 θM = cos−1 √ cos θz 3

(7.6.16)

Since the maximum power loss ratio must be equal to 1 + k 2 , k2 =

(ZL − Z0 )2 4ZL Z0

2 cos θz √ 3 3 tan2 θz

2 (7.6.17)

Characteristic impedance Z1 is obtained by solving the equation 1/2 1/2 ZL Z02 Z12 ZL ZL ZL − Z0 = +2 Z1 − −2 Z1−1 Z02 tan2 θz Z0 Z0 Z0 Z12

(7.6.18)

The other two characteristic impedances, Z2 and Z3 , are then determined as follows: Z2 = ZL Z0 (7.6.19) and Z3 =

ZL Z0 Z1

(7.6.20)

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

261

The range of the passband (bandwidth) is still given by (7.6.11) provided that θM is now computed from (7.6.16). Example 7.6 Design a three-section Chebyshev transformer (exact theory) to match a 500- load to a 100- line. The required fractional bandwidth is 0.6. Compute ρM and compare it with that obtained in Example 7.5. SOLUTION From (7.6.16) and (7.6.11), (π/2) − θM 2 f = 0.6 ⇒ and √ cos θz f0 π/2 3 π = 0.3 ⇒ θM = (1 − 0.3) = 1.1 2

θM = cos−1

Therefore,

√ cos θz =

3 cos θM = 0.3932 ⇒ θz = 1.17 rad 2

Now, from (7.6.17), (ZL − Z0 )2 k = 4ZL Z0 2

2 cos θz √ 3 3 tan2 θz

2

and from (7.6.4),

(ZL − 1)2 = 4ZL

2 cos θz √ 3 3 tan2 θz

2

= 6.125 × 10−4

ρM =

k2 = 0.0247 1 + k2

which is approximately one-fourth of the previous (two-section transformer) case. Z1 is obtained by solving equation (7.6.18) as follows: 1/2 1/2 ZL Z02 ZL − Z0 Z12 ZL ZL = + 2 Z − − 2 Z1−1 Z02 1 tan2 θz Z0 Z0 Z0 Z12 or 4 ZL − 1 ZL 2 −1 = Z 1 + 2(Z L )1/2 Z 1 − 2 − 2(Z L )1/2 Z 1 = = 0.7314 2θ tan2 θz tan z Z1 Therefore, 2 Z1

or

√ √ 5 2 5 + 2 5 Z1 − 2 − = 0.7314 Z1 Z1

√ √ 4 3 2 Z 1 + 2 5 Z 1 − 0.7314 Z 1 − 2 5 Z 1 − 5 = 0

262

IMPEDANCE TRANSFORMERS

Therefore, Z 1 = −4.4679, (−0.6392 − j 0.6854), (−0.6392 + j 0.6854), and 1.2741 Thus, only one of these solutions can be physically realized because the others have a negative real part. After selecting Z1 = 127.42 , (7.6.19) and (7.6.20) give Z3 =

ZL Z0 = 392.4277 Z1

Z2 =

ZL Z0 = 223.6068

and

If we use the approximate formulation, then following the procedure of Example 7.4, we find from (7.5.6) that T3 (x0 ) =

500 − 100 1 × = 26.9458 500 + 100 0.02474

Now, from (7.5.7), x0 = cosh[ 31 cosh−1 (26.9458)] = 2.0208 and from (7.5.8),

π = ±0.866, 0 xn = ± cos (2n − 1) 6 Using (7.5.9) we get ϕn = 2.2558, 3.1416, and 4.0274 Since wn = ej ϕn , we find that w1 = −0.6327 + j 0.7744 w2 = −0.6327 − j 0.7744 w3 = −1 and from (7.2.9), = (w − w1 )(w − w2 )(w − w3 ) = w3 + 2.2654w2 + 2.2654w + 1 3 Hence, 0 = 3 1 = 2 = 2.26543

TAPERED TRANSMISSION LINES

263

Now, from (7.2.12), 0 + 1 + 2 + 3 = 6.53073 = Therefore,

500 − 100 500 + 100

3 = 0.1021 = 0 1 = 2 = 0.2313

Corresponding impedances are now determined from (7.2.8) as follows: 1 + 0 Z1 = = 1.2274 1 − 0 calculated from the input side 1 + 1 Z2 = Z 1 = 2.4234 1 − 1 and 1 − 3 Z3 = Z L = 4.0737 calculated from the load side 1 + 3 A comparison of these values with those obtained earlier using the exact formulation shows that the two sets are within 10% of deviation for this example. Note that for the fractional bandwidth of 0.6 with a single section, reflection coefficient ρM increases to 0.3762. 7.7 TAPERED TRANSMISSION LINES Consider that the multisection impedance transformer of Figure 7.4 is replaced by a tapered transition of length L, as illustrated in Figure 7.21. The characteristic impedance of this transition is a continuous smooth function of distance, with its values at the two ends as Z0 and RL . An approximate theory of such a matching section can easily be developed on the basis of the analysis presented in Section 7.2. The continuously tapered line can be modeled by a large number of incremental sections of length δz. One of these sections, connected at z, has a characteristic impedance of Z + δZ, and the one before it has a characteristic impedance of Z, as shown in Figure 7.21. These impedance values are conveniently normalized by Z0 before obtaining the incremental reflection coefficient at distance z. Hence, δ0 =

Z + δZ − Z Z + δZ + Z

≈

δZ 2Z

(7.7.1)

As δz → 0, it can be written as d0 =

dZ 2Z

=

1 d(ln Z) dz 2 dz

(7.7.2)

264

IMPEDANCE TRANSFORMERS

L

δz

Z Z0

RL

Z + δZ

z

Figure 7.21 Tapered transition connected between load RL and transmission line of characteristic impedance Z0 .

The corresponding incremental reflection coefficient din at the input can be written as follows: d −j 2βz −j 2βz 1 d0 = e (ln Z) dz din ≈ e 2 dz The total reflection coefficient, in , at the input of the tapered section can be determined by summing up these incremental reflections with their appropriate phase angles. Hence,

L

in = 0

din =

1 2

L

e−j 2βz

0

d (ln Z) dz dz

(7.7.3)

Therefore, in can be determined from (7.7.3) provided that Z(z) is given. However, the synthesis problem is a bit complex because in that case, Z(z) is to be determined for a specified in . Let us first consider a few examples of evaluating in for the given distributions of d(ln Z)/dz. Case 1

d (ln Z) is constant over the entire length of the taper. Suppose that dz d ln Z(z) = C1 dz

0≤z≤L

where C1 is a constant. On integrating this equation, we get ln Z(z) = C1 z + C2 With Z(z = 0) = 1 and Z(z = L) = R L , constants C1 and C2 can be determined. Hence, z ln Z = ln R L ⇒ Z = e(z/L) ln RL (7.7.4) L

265

TAPERED TRANSMISSION LINES

Thus, the impedance is changing exponentially with distance. Therefore, this kind of taper is called an exponential taper. From (7.7.3), the total reflection coefficient is found as L ln R L L −j 2βz 1 L −j 2βz d z ln R L e−j 2βz in = e e dz = ln R L dz = 2 0 dz L 2L 0 2L −j 2β 0 or in =

ln R L e−j βL − 1 sin βL ln R L −j βL sin βL 2ρin = = e ⇒ 2L −j 2β 2 βL βL ln R L

(7.7.5)

where ρin = |in |. It is assumed in this evaluation that β = 2π/λ is not changing with distance z. The right-hand side of (7.7.5) is displayed in Figure 7.22 as a function of βL. Since L is fixed for a given taper and β is related directly to signal frequency, this plot also represents the frequency response of an exponential taper. d d (ln Z) is a triangular function. If (ln Z) is a triangular function, dz dz defined as 4z L ln R L 0≤z≤ 2 d(ln Z) L2 = (7.7.6) 1 L z dz 4 ln R L − ≤z≤L L L2 2 Case 2

1.0

0.8

0.6

0.4

0.2

0.0 0.0000

3.1416

6.2832

9.4248 βL

12.5664

15.7080

18.8496

Figure 7.22 Frequency response of an exponential taper.

21.9912

266

then

or

IMPEDANCE TRANSFORMERS

L/2 4z ln R L dz 0 L2 ln Z = 1 z L ln R L dz − L/2 4 L L2 2 2z 2 ln R L + c1 L ln Z = z z2 4 ln R L + c2 − L 2L2

0≤z≤

L 2

L ≤z≤L 2 0≤z≤

L 2

L ≤z≤L 2

Integration constants c1 and c2 are determined from the given conditions Z(z = 0) = 1 and Z(z = L) = R L . Hence, 2 L 2z 0≤z≤ exp L2 ln R L 2 (7.7.7) Z= 2 4z 2z L − ≤ z ≤ L exp − 1 ln R L L L2 2 and therefore total reflection coefficient is found from (7.7.3) as in =

1 2

L/2 0

or in =

e−j 2βz

2 ln R L L2

L/2

4z ln R L dz + L2

ze−j 2βz dz +

0

L L/2

L

e−j 2βz

4 (L − z) ln R L dz L2

(L − z)e−j 2βz dz

L/2

Since

and

ze−j 2βz dz =

e−j 2βz dz =

ze−j 2βz e−j 2βz − −j 2β (−j 2β)2 e−j 2βz −j 2β

in can be found as follows: in =

1 −j βL sin(βL/2) 2 ln R L e 2 βL/2

(7.7.8)

Figure 7.23 shows normalized magnitude of in as a function of βL. As before, it also represents the frequency response of this taper of length L. When we compare it with that of Figure 7.22 for an exponential taper, we find that the

267

TAPERED TRANSMISSION LINES 1.0

0.8

0.6

0.4

0.2

0.0 0.0000

3.1416

6.2832

9.4248

12.5664

15.7080

18.8496

21.9912

βL

Figure 7.23 Frequency response of a taper with triangular distribution.

magnitude of minor lobes is reduced now, but the main lobe has become relatively wider. It affects the cutoff frequency of a given taper. For example, if the maximum allowed normalized reflection coefficient is 0.2, an exponential taper has a lower value of βL than that of a taper with triangular distribution. However, it is the other way around for a normalized reflection coefficient of 0.05. For a fixed length L, βL is proportional to frequency, and hence the lower value of βL has a lower cutoff frequency than in the other case. Figure 7.24 depicts variation in normalized characteristic impedance along the length of two tapers. A coaxial line operating in the TEM mode can be used to design these tapers by varying the diameter of its inner conductor. Similarly, the narrow sidewall width of a TE10 -mode rectangular waveguide can be modified to get the desired taper while keeping its wide side constant. Case 3

d (ln Z) is a Gaussian distribution. Assume that dz d 2 (ln Z) = K1 e−α(z−L/2) dz

(7.7.9)

This distribution is centered around the midpoint of the tapered section. Its fall-off rate is governed by the coefficient α. Integrating it over distance z, we find that

z

ln Z = K1 0

e−α(z −L/2) dz 2

(7.7.10)

268

IMPEDANCE TRANSFORMERS 2.0

1.8 (b)

(a)

1.6

1.4

1.2

1.0 0.0

0.2

0.4

0.6

0.8

1.0

Figure 7.24 Distribution of normalized characteristic impedance along a tapered line terminated by RL = 2Z0 for (a) case 1 and (b) case 2.

Since Z(z = L) = R L , ln R L

K1 =

L

e

−α(z −L/2)2

(7.7.11) dz

0

Therefore,

z

e−α(z −L/2) dz

ln Z = ln R L 0L

2

(7.7.12) e

−α(z −L/2)2

dz

0

and in =

1 K1 2

L

e−j 2βz e−α(z−L/2) dz 2

(7.7.13)

0

With the following substitution of variable and associated limits of integration in (7.7.11) and (7.7.13), z−

L =x 2 dz = dx z=0→x=−

L 2

269

TAPERED TRANSMISSION LINES

and z=L→x=

L 2

we get in =

1 K1 2

L/2

−L/2

1 K1 e−j βL 2

e−j 2β(x+L/2) e−αx dx =

or

2

1 in = ln R L e−j βl 2

L/2

L/2

e−j 2βx e−αx dx

−L/2

2

L/2

e

−αx 2

2

e−j 2βx e−αx dx

1 = ln R L e−j βl 2

−L/2

dx

L/2

0

cos 2βx e−αx dx 2

−L/2

L/2

e−αx dx 2

0

(7.7.14)

It can be arranged after substituting

√

α x with y, and

1 in = ln R L e−j βl 2

ξ 0

√

α(L/2) with ξ as follows:

cos[(βL/ξ)y]e−y dy ξ 2 e−y dy 2

(7.7.15)

0

Figure 7.25 shows the normalized magnitude of the reflection coefficient versus βL for two different values of ξ. It indicates that the main lobe becomes wider, and the level of minor lobes goes down as ξ is increased.

1 0.8 (a) 0.6 0.4

(b)

0.2 0 0

5

10

15

20

βL

Figure 7.25 Frequency response of a tapered line with a normal distribution of normalized characteristic impedance for (a) ξ = 0.3 and (b) ξ = 3.

270

IMPEDANCE TRANSFORMERS

7.8 SYNTHESIS OF TRANSMISSION LINE TAPERS The frequency response of a transmission line taper can be determined easily from (7.7.3). However, the inverse problem of determining its impedance variation that provides the desired frequency response needs further consideration. To this end, assume that F (α) is the Fourier transform of a function f (x). In other words, ∞ f (x)e−j αx dx (7.8.1) F (α) = −∞

and

1 f (x) = 2π

∞

F (α)ej αx dα

(7.8.2)

−∞

For convenience, we rewrite (7.7.3) as follows: L 1 d in (2β) = (ln Z) e−j 2βz dz 2 dz 0 If

1 2

d(ln Z)/dz = 0 for −∞ ≤ z < 0 and for z > L, it may be written as ∞ 1 d (7.8.3) in (2β) = (ln Z)e−j 2βz dz −∞ 2 dz

A comparison of (7.8.3) with (7.8.1) indicates that in (2β) represents the Fourier transform of 12 d[ln Z(z)]/dz. Hence, from (7.8.2), ∞ 1 1 d(ln Z) = in (2β)ej 2βx d(2β) (7.8.4) 2 dz 2π −∞ This equation can be used to design a taper that will have the desired reflection coefficient characteristics. However, only those in (2β) that have an inverse Fourier transform limited to 0 ≤ z ≤ L (zero outside this range) can be realized. At this point, we can conveniently introduce the following normalized variables: p = 2π

z − (L/2) L

(7.8.5)

and u=

2L βL = π λ

Therefore, z=

L (p + π) 2π

and dz =

L dp 2π

(7.8.6)

SYNTHESIS OF TRANSMISSION LINE TAPERS

New integration limits are found as −π ≤ p ≤ π. Hence, 1 π −j 2β(L/2+Lp/2π) d dp L (ln Z) dp in (2β) = e 2 −π dp dz 2π or π 1 d in (2β) = e−j βL e−j up (ln Z) dp 2 dp −π Now, if we define g(p) = d(ln Z)/dp and F (u) = can be expressed as follows:

π

−π

271

(7.8.7)

e−j up g(p) dp, (7.8.7)

in (2β) = 12 e−j βL F (u) or ρin (2β) = 12 |F (u)| Further, F (u = 0) = ln Z L

(7.8.8)

Thus, F (u) and g(p) form the Fourier transform pair. Therefore, ∞ 1 nonzero for |p| < π jup e F (u) du = g(p) = 0 for |p| > π 2π −∞

(7.8.9)

To satisfy the conditions embedded in writing (7.8.3), only those F (u) that produce g(p) as zero for |p| > π can be synthesized. Suitable restrictions on F (u) can be derived after expanding g(p) in a complex Fourier series as follows: ∞ a ejnp −π ≤ p ≤ π n g(p) = n=−∞ (7.8.10) 0 |p| > π ∗ Since g(p) is a real function, constants an = a−n . Therefore,

F (u) =

π

e

−j up

−π

∞

an e

n=−∞

jnp

dp =

∞

an

n=−∞

π

e −π

−j (u−n)p

π e−j (u−n)p dp = an −j (u − n) −π n=−∞ ∞

or F (u) =

∞ n=−∞

an

∞ e−j (u−n)π − ej (u−n)π sin π(u − n) = 2π an −j (u − n) π(u − n) n=−∞

For u = m (an integer), sin π(m − n) = π(m − n)

1 for m = n 0 for m = n

(7.8.11)

272

IMPEDANCE TRANSFORMERS

Therefore, F (n) = 2πan ⇒ an =

F (n) 2π

and F (u) =

∞

F (n)

n=−∞

sin π(u − n) π(u − n)

(7.8.12)

This is the well-known sampling theorem used in communication theory. It states that F (u) is uniquely reconstructed from a knowledge of sample values of F (u) at u = n, where n is an integer (positive or negative) including zero (i.e., u = n = 0, ±1, ±2, ±3, . . .). To have greater flexibility in selecting F (u), let us assume that an = 0 for all |n| > N . Therefore, N an ejnp (7.8.13) g(p) = n=−N

and from (7.8.11), F (u) = 2π

N n=−N

an (−1)n

N sin πu u sin πu an (−1)n = 2π π(u − n) πu n=−N u−n

This series can be recognized as the partial-fraction expansion of a function, and therefore F (u) can be expressed as follows: F (u) = 2π

sin πu Q(u) N 2 2 πu n=1 (u − n )

(7.8.14)

where Q(u) is an arbitrary polynomial of degree 2N in u subject to the restriction ∗ . Further, it contains an arbitrary constant mulQ(−u) = Q∗ (u), so that an = a−n tiplier. In (7.8.14), sin πu has zeros at u = ±n. However, the first 2N of these zeros are canceled by (u2 − n2 ). These can be replaced by 2N new arbitrarily located zeros by proper choice of Q(u). Example 7.7 Design a transmission line taper to match a 100- load to a 50 line. The desired frequency response has a triple zero at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution. SOLUTION The desired characteristic can be accomplished by moving zeros at ±1 and ±3 into the points u = ±2. Due to this triple zero, the reflection coefficient will remain small over a relatively wide range of βL around ±2π. Therefore, Q(u) = C(u2 − 4)3

SYNTHESIS OF TRANSMISSION LINE TAPERS

and F (u) = 2πC

273

sin πu (u2 − 4)3 πu (u2 − 1) × (u2 − 4) × (u2 − 9)

Since u = βL/π, F (u) versus u represents the frequency response of this taper. Further, it can be normalized with F (0) as follows: F (u) (u2 − 4)3 = 9 sin πu F (0) 16 πu (u2 − 1)(u2 − 4)(u2 − 9) A plot of this equation is illustrated in Figure 7.26. It shows that the normalized magnitude is very small around u = 2. Since F (0) is given by (7.8.8), multiplying constant C in F (u) can be determined easily. Hence, F (0) = 2πC × 1 ×

(−4)3 9 16 = 2πC × = ln Z L ⇒ C = ln Z L (−1)(−4)(−9) 9 32π

Further, 2πC F (1) = π

(1 − 4)3 sin πu 9 × = Cπ u2 − 1 u=1 (1 − 4)(1 − 9) 8

9 9 ln Z L π = 0.3164 ln Z L 32π 8 F (2) = 0 25 (9 − 4)3 2πC sin πu = − Cπ × F (3) = 3π u2 − 9 u=3 (9 − 1)(9 − 4) 72 25 9 = − π × ln Z L = 0.09766 ln Z L 72 32π =

and F (n) = 0 for n > 3. Therefore, 1 1 F (0) = ln Z L 2π 2π 1 0.3164 a1 = a−1 = F (1) = ln Z L 2π 2π a2 = a−2 = 0

a0 =

a3 = a−3 =

0.09766 1 F (3) = − ln Z L 2π 2π

and an = a−n = 0

for n > 3

274

IMPEDANCE TRANSFORMERS 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

u

Figure 7.26 Normalized frequency response of the taper in Example 7.7.

Hence, g(p) =

d(ln Z) ln Z L = (a0 + 2a1 cos p + 2a3 cos 3p) dp 2π

=

ln Z L (1 + 0.6328 × cos p − 0.1953 × cos 3p) 2π

and ln Z =

ln Z L (p + 0.6328 × sin p − 0.0651 × sin 3p) + c1 2π

The integration constant c1 can be evaluated as follows: ln Z = 0 at p = −π or

ln Z = ln Z L at p = π

Therefore, c1 = 0.5 ln Z L and ln Z = where

ln Z L (p + π + 0.6328 sin p − 0.0651 sin 3p) 2π L 2π z− p= L 2

Normalized impedance (Z) versus the length of the taper (z/L) is illustrated in Figure 7.27.

275

SYNTHESIS OF TRANSMISSION LINE TAPERS 2.0

Normalized impedance

1.8 1.6 1.4 1.2 1.0 0.8 0.0

0.2

0.4

0.6

0.8

1.0

z/L

Figure 7.27 Normalized characteristic impedance variation along the length of the taper in Example 7.7.

Example 7.8 Design a taper to match a 100- load to a 50- transmission line. Its reflection coefficient has double zeros at βL = ±2π and ±3π. Plot its frequency response and normalized characteristic impedance versus the normalized length of the taper. SOLUTION This design can be achieved by moving the zeros at u = ± 1 and ± 4 into the points ± 2 and ± 3, respectively. Hence, Q(u) = C(u2 − 22 )2 (u2 − 32 )2 = C(u2 − 4)2 (u2 − 9)2 and F (u) = C

sin πu (u2 − 4)2 (u2 − 9)2 πu (u2 − 1)(u2 − 4)(u2 − 9)(u2 − 16)

Therefore, F (0) = 2.25C, F (1) = 0.8C, F (2) = 0, F (3) = 0, F (4) = (7/40)C, and F (n) = 0 for n > 4. From (7.8.8), F (0) = ln Z L ⇒ C = and g(p) =

∞ d(ln Z) an e−jnp = dp n=−∞

ln Z L 2.25 an =

F (n) 2π

or g(p) = =

2.25C 7C 0.8C jp + (e + e−jp ) + (ej 4p + e−j 4p ) 2π 2π (40)(2π) ln Z L (1 + 0.7111 cos p + 0.1556 cos 4p) 2π

276

IMPEDANCE TRANSFORMERS 1 0.8 0.6 0.4 0.2 0 0

2

4

6

8

u

Figure 7.28 Frequency response of the taper in Example 7.8.

Therefore, ln Z =

ln Z L (p + 0.7111 sin p + 0.0389 sin 4p) + c1 2π

Since, ln Z|p=−π = 0, c1 = ln Z L /2 and ln Z =

ln Z L (p + π + 0.7111 sin p + 0.0389 sin 4p) 2π

where p=

2π L z− L 2

Frequency response and normalized characteristic impedance distribution are shown in Figures 7.28 and 7.29, respectively.

Klopfenstein Taper The preceding procedure indicates infinite possibilities of synthesizing a transmission line taper. A designer naturally looks for the best design. In other words, which design gives the shortest taper for a given ρM ? The answer to this question is the Klopfenstein taper, which is derived from a stepped Chebyshev transformer. The design procedure is summarized here without its derivation. The characteristic impedance variation of this taper versus distance z is given by Z(z) =

Z0 RL eκ

(7.8.15)

277

SYNTHESIS OF TRANSMISSION LINE TAPERS 2.0

Normalized impedance

1.8

1.6

1.4

1.2

1.0

0.8 0.0

0.2

0.4

0.6

0.8

1.0

z /L

Figure 7.29 Normalized characteristic impedance variation of the taper in Example 7.8.

where

A2 0 2z f − 1, A cosh A L ζ I1 (A 1 − y 2 ) dy f (ζ, A) = A 1 − y2 0 κ=

0 =

0≤z≤L |x| ≤ 1

RL − Z0 RL + Z0

(7.8.16) (7.8.17) (7.8.18)

Since the cutoff value of βL is equal to A, we have A = β0 L

(7.8.19)

For a maximum ripple ρM in its passband, it can be determined as A = cosh−1

|0 | ρM

(7.8.20)

I1 (x) is the modified Bessel function of the first kind. f (ζ, A) needs to be evaluated numerically except for the following special cases: f (0, A) = 0 x f (x, 0) = 2 cosh A − 1 f (1, A) = A2

278

IMPEDANCE TRANSFORMERS

The resulting input reflection coefficient as a function of βL (and hence, frequency) is given by in (βL) = 0 e

−j βL cosh

A2 − (βL)2 cosh A

(7.8.21)

Note that the hyperbolic cosine becomes the cosine function for an imaginary argument. Example 7.9 Design a Klopfenstein taper to match a 100- load to a 50- transmission line. The maximum allowed reflection coefficient in its passband is 0.1. Plot the characteristic impedance variation along its normalized length and the input reflection coefficient versus βL. SOLUTION From (7.8.18), 0 =

RL − Z0 100 − 50 1 = = RL + Z0 100 + 50 3

TABLE 7.1 Characteristic Impedance of the Klopfenstein Taper (in Example 7.9) as a Function of Its Normalized Length Z/L

Impedance ()

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

56.000896 57.030043 58.169255 59.412337 60.757724 62.202518 63.74238 65.371433 67.082194 68.865533 70.710678 72.605261 74.535428 76.486009 78.440749 80.382597 82.294063 84.157604 85.95606 87.673089 89.29361

279

SYNTHESIS OF TRANSMISSION LINE TAPERS 90.0

Characteristic impedance (Ω)

85.0 80.0 75.0 70.0 65.0 60.0 55.0 0.0

0.2

0.4

0.6

0.8

1.0

z/L

Figure 7.30 Characteristic impedance distribution along the normalized length of the Klopfenstein taper in Example 7.9.

0.3 0.25

|Γin(βL)|

0.2 0.15 0.1 0.05 0 0

5

10

15

20

βL

Figure 7.31 Reflection coefficient versus βL of the Klopfenstein taper in Example 7.9.

From (7.8.20), A = cosh−1

1 |0 | = cosh−1 3 = 1.87382 ρM 0.1

Equation (7.8.15) is used to evaluate Z(z) numerically. These results are shown in Table 7.1. These data are also displayed in Figure 7.30. The reflection coefficient of this taper is depicted in Figure 7.31.

280

IMPEDANCE TRANSFORMERS

7.9 BODE–FANO CONSTRAINTS FOR LOSSLESS MATCHING NETWORKS Various matching circuits described in this and Chapter 6 indicate that zero reflection is possible only at selected discrete frequencies. Examples 6.6 and 6.7 show that the bandwidth over which the reflection coefficient remains below a specified value is increased with the number of components. A circuit designer would like to know whether a lossless passive network could be designed for perfect matching. Further, it will be helpful if associated constraints and design trade-offs are known. Bode–Fano constraints provide such means to the designer. Bode–Fano criteria provide optimum results under ideal conditions. These results may require approximation to implement the circuit. Table 7.2 summarizes these constraints for selected R –C and R –L loads. The detailed formulations are beyond the scope of this book. As an example, if it is desired to design a reactive matching network at the output of a transistor amplifier, the situation is similar to that depicted in the top row of the table. Further, consider the case in which magnitude ρ of the reflection TABLE 7.2 Bode–Fano Constraints to Match Certain Loads Using Passive Lossless Network Circuit Arrangement

Constraint Relation

Γ = ρ∠θ

Matching network

0

R

Γ = ρ∠θ

∞

C

Matching network

L

R

∞

0

π 1 ln dω ≤ ρ RC

ω2 πL 1 ln dω ≤ 0 ρ R

(where ω0 is the center frequency) Γ = ρ∠θ

Matching network

∞

C R

0

1 ln dω ≤ ω20 πRC ρ

(where ω0 is the center frequency)

Γ = ρ∠θ

Matching network

L

R 0

∞

πR 1 ln dω ≤ ρ L

PROBLEMS

281

coefficient remains constant at ρm (ρm < 1) over the frequency band from ω1 to ω2 , whereas it stays at unity outside this band. The constraint for this circuit is found to be ∞ ω2 1 1 1 π (7.9.1) ln dω = ln dω = (ω2 − ω1 ) ln ≤ ρ ρ ρ RC m ω1 m 0 This condition shows the trade-off associated with the bandwidth and the reflection coefficient. If R and C are given, the right-hand side of (7.9.1) is fixed. In this situation, a wider bandwidth (ω2 − ω1 ) is possible only at the cost of a higher reflection coefficient ρm . Further, ρm cannot go to zero unless the bandwidth ω2 − ω1 is zero. In other words, a perfect matching condition is achievable only at discrete frequencies.

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures. Dedham, MA: Artech House, 1980. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998.

PROBLEMS 7.1. The output impedance of a microwave receiver is 50 . Find the required length and characteristic impedance of a transmission line that will match a 100- load to the receiver. Signal frequency and phase velocity are 10 GHz and 2.4 × 108 m/s, respectively. What is the frequency band over which the reflection coefficient remains below 0.1? 7.2. A quarter-wave impedance-matching transformer is used to match a microwave source with internal impedance 10 to a 50- transmission line. If the source frequency is 3 GHz and the phase velocity on the line is equal to the speed of light in free space, find (a) the required length and characteristic impedance of the matching section and (b) the reflection coefficients at two ends of the matching section. 7.3. A 0.7-m-long transmission line short-circuited at one end has an input impedance of −j 68.8 . The signal frequency and phase velocity are 100 MHz and 2 × 108 m/s, respectively.

282

IMPEDANCE TRANSFORMERS

(a) What is the characteristic impedance of the transmission line? (b) If a 200- load is connected to the end of this line, what is the new input impedance? (c) A quarter-wave matching transformer is to be used to match a 200- load with the line. What will be the required length and characteristic impedance of the matching transformer? 7.4. Design a two-section binomial transformer to match a 100- load to a 75 line. What is the frequency band over which the reflection coefficient remains below 0.1? 7.5. Design a three-section binomial transformer to match a 75- load to a 50- transmission line. What is the fractional bandwidth for a VSWR less than 1.1? 7.6. Using the approximate theory, design a two-section Chebyshev transformer to match a 100- load to a 75- line. What is the frequency band over which the reflection coefficient remains below 0.1? Compare the results with those obtained in Problem 7.4. 7.7. Design a four-section Chebyshev transformer to match a network with input impedance 60 to a transmission line of characteristic impedance 40 . Find the bandwidth if the maximum VSWR allowed in its passband is 1.2. 7.8. Using the exact theory, design a two-section Chebyshev transformer to match a 10- load to a 75- line. Find the bandwidth over which the reflection coefficient remains below 0.05. 7.9. Using the exact theory, design a three-section Chebyshev transformer to match a 10- load to a 75- line. Find the bandwidth over which the reflection coefficient remains below 0.05. Compare the results with those obtained in Problem 7.8. 7.10. Design a transmission line taper to match a 75- load to a 50- line. The desired frequency response has a double zero at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution. 7.11. Design a transmission line taper to match a 40- load to a 75- line. Its reflection coefficient has triple zeros at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution versus the normalized length of the taper. 7.12. Design a Klopfenstein taper to match a 40- load to a 75- line. The maximum allowed reflection coefficient in its passband is 0.1. Plot the characteristic impedance variation along its normalized length and the input reflection coefficient versus βL. Compare the results with those obtained in Problem 7.11.

8 TWO-PORT NETWORKS

Electronic circuits are frequently needed for processing a given electrical signal to extract the desired information or characteristics. This includes boosting the strength of a weak signal or filtering out certain frequency bands, and so on. Most of these circuits can be modeled as a black box that contains a linear network comprising resistors, inductors, capacitors, and dependent sources. Thus, it may include electronic devices but not the independent sources. Further, it has four terminals, two for input and the other two for output of the signal. There may be a few more terminals to supply the bias voltage for electronic devices. However, these bias conditions are embedded in equivalent dependent sources. Hence, a large class of electronic circuits can be modeled as two-port networks. Parameters of the two-port completely describe its behavior in terms of voltage and current at each port. These parameters simplify the description of its operation when the two-port network is connected into a larger system. Figure 8.1 shows a two-port network along with appropriate voltages and currents at its terminals. Sometimes, port 1 is called the input port and port 2, the output port. The upper terminal is customarily assumed to be positive with respect to the lower one on either side. Further, currents enter the positive terminals at each port. Since the linear network does not contain independent sources, the same currents leave respective negative terminals. There are several ways to characterize this network. Some of these parameters and relations among them are presented in this chapter, including impedance parameters, admittance parameters, hybrid

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

283

284

TWO-PORT NETWORKS I1

I2 +

+ V1

Port 1

Linear network

Port 2

V2 −

−

Figure 8.1

Two-port network.

parameters, and transmission parameters. Scattering parameters are introduced later in the chapter to characterize the high-frequency and microwave circuits. 8.1 IMPEDANCE PARAMETERS Consider the two-port network shown in Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 at port 1 can be expressed in terms of two currents as follows: V1 = Z11 I1 + Z12 I2 (8.1.1) Since V1 is in volts and I1 and I2 are in amperes, parameters Z11 and Z12 must be in ohms. Therefore, these are called the impedance parameters. Similarly, we can write V2 in terms of I1 and I2 as follows: V2 = Z21 I1 + Z22 I2 Using the matrix representation, we can write Z11 I1 V1 Z12 = V2 Z21 Z22 I2

(8.1.2)

(8.1.3)

or [V ] = [Z][I ]

(8.1.4)

where [Z] is called the impedance matrix of the two-port network. If port 2 of this network is left open, I2 will be zero. In this condition, (8.1.1) and (8.1.2) give V1 Z11 = (8.1.5) I1 I2 =0 and Z21

V2 = I1 I2 =0

(8.1.6)

Similarly, with a source connected at port 2 while port 1 is open circuit, we find that V1 Z12 = (8.1.7) I2 I1 =0

285

IMPEDANCE PARAMETERS

and Z22 =

V2 I2 I1 =0

(8.1.8)

Equations (8.1.5) through (8.1.8) define the impedance parameters of a twoport network. Example 8.1 Find the impedance parameters for the two-port network shown in Figure 8.2. SOLUTION If I2 is zero, V1 and V2 can be found from Ohm’s law as 6I1 . Hence, from (8.1.5) and (8.1.6), V1 6I1 Z11 = = = 6 I1 I2 =0 I1 and Z21 =

V2 6I1 = = 6 I1 I2 =0 I1

Similarly, when a source is connected at port 2 and port 1 has an open circuit, we find that V2 = V1 = 6I2 Hence, from (8.1.7) and (8.1.8), Z12 = and Z22 Therefore,

V1 6I2 = = 6 I2 I1 =0 I2

V2 6I2 = = = 6 I2 I1 =0 I2

Z11 Z21

Z12 Z22

6 = 6

6 6

I1

I2 + V1 −

Figure 8.2

+ 6Ω

V2 −

Two-port network for Example 8.1.

286

TWO-PORT NETWORKS I1

+

+

V1

I2

V2

12 Ω

3Ω −

−

Figure 8.3

Two-port network for Example 8.2.

Example 8.2 Find the impedance parameters of the two-port network shown in Figure 8.3. SOLUTION As before, assume that a source is connected at port 1 while port 2 is open. In this condition, V1 = 12I1 and V2 = 0. Therefore, V1 12I1 Z11 = = = 12 I1 I2 =0 I1 and Z21

V2 = =0 I1 I2 =0

Similarly, when a source connected at port 2 and port 1 has an open circuit, we find that V2 = 3I2 and V1 = 0 Hence, from (8.1.7) and (8.1.8), Z12 and Z22 Therefore,

Z11 Z21

V1 = =0 I2 I1 =0

V2 3I2 = = = 3 I2 I1 =0 I2 Z12 Z22

12 = 0

0 3

Example 8.3 Find the impedance parameters for the two-port network shown in Figure 8.4. SOLUTION Assuming that the source is connected at port 1 while port 2 is open, we find that V1 = (12 + 6)I1 = 18I1

and V2 = 6I1

287

IMPEDANCE PARAMETERS 12 Ω I1

3Ω

+

+

V1

I2

V2

6Ω

−

−

Figure 8.4

Two-port network for Example 8.3.

Note that there is no current flowing through a 3- resistor because port 2 is open. Therefore, V1 18I1 = = 18 Z11 = I1 I2 =0 I1 and Z21

V2 6I1 = = = 6 I1 I2 =0 I1

Similarly, with a source at port 2 and port 1 open circuit, V2 = (6 + 3)I2 = 9I2

and V1 = 6I2

This time, there is no current flowing through a 12- resistor because port 1 is open. Hence, from (8.1.7) and (8.1.8), V1 6I2 Z12 = = = 6 I2 I1 =0 I2 and Z22 Therefore,

Z11 Z21

V2 9I2 = = = 9 I2 I1 =0 I2 Z12 Z22

=

18 6

6 9

An analysis of results obtained in Examples 8.1 to 8.3 indicates that Z12 and Z21 are equal for all three circuits. In fact, it is an inherent characteristic of these networks. It will hold for any reciprocal circuit. If a given circuit is symmetrical, Z11 will be equal to Z22 as well. Further, impedance parameters obtained in Example 8.3 are equal to the sum of the corresponding results found in Examples 8.1 and 8.2. This happens because if the circuits of these two examples are connected in series, we end up with the circuit of Example 8.3, shown in Figure 8.5.

288

TWO-PORT NETWORKS I1

I2 +

+ V1

3Ω 12 Ω

−

V2 −

I1

I2 +

+ 6Ω

V ′1

V ′2 −

−

Figure 8.5 Circuit for Example 8.3.

Example 8.4 Find the impedance parameters for a transmission line network shown in Figure 8.6. SOLUTION This circuit is symmetrical because interchanging port 1 and port 2 does not affect it. Therefore, Z22 must be equal to Z11 . Further, if current I at port 1 produces an open-circuit voltage V at port 2, current I injected at port 2 will produce V at port 1. Hence, it is a reciprocal circuit. Therefore, Z12 will be equal to Z21 . Assume that the source is connected at port 1 while the other port is open. If Vin is incident voltage at port 1, Vin e−γl is the voltage at port 2. Since the reflection coefficient of an open circuit is +1, the reflected voltage at this port is equal to the incident voltage. Therefore, the reflected voltage reaching port 1 is Vin e−2γl . Hence, V1 = Vin + Vin e−2γl V2 = 2Vin e−γl Vin I1 = (1 − e2γl ) Z0 l I1

+

V1

−

+ Z0, γ

V2

−

Figure 8.6 Transmission line network for Example 8.4.

I2

ADMITTANCE PARAMETERS

289

and I2 = 0 Therefore, Z11 =

V1 Vin (1 + e−2γl ) Z0 e+γl + e−γl = Z0 +γl = Z0 coth γl = = −2γl I1 I2 =0 (Vin /Z0 )(1 − e ) e − e−γl tanh γl

and Z21

V2 2Vin e−γl Z0 2 = = = = Z0 +γl −2γl −γl I1 I2 =0 (Vin /Z0 )(1 − e ) e −e sinh γl

For a lossless line, γ = j β, and therefore, Z11 =

Z0 = −j Z0 cot βl j tan βl

and Z21 =

Z0 Z0 = −j j sin βl sin βl

8.2 ADMITTANCE PARAMETERS Consider again the two-port network shown in Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, current I1 at port 1 can be expressed in terms of two voltages: I1 = Y11 V1 + Y12 V2 (8.2.1) Since I1 is in amperes and V1 and V2 are in volts, parameters Y11 and Y12 must be in siemens. Therefore, these are called the admittance parameters. Similarly, we can write I2 in terms of V1 and V2 as follows: I2 = Y21 V1 + Y22 V2 Using the matrix representation, we can write I1 Y11 Y12 V1 = I2 Y21 Y22 V2 or [I ] = [Y ][V ] where [Y ] is called the admittance matrix of the two-port network.

(8.2.2)

(8.2.3)

(8.2.4)

290

TWO-PORT NETWORKS

If port 2 of this network has a short circuit, V2 will be zero. In this condition, (8.2.1) and (8.2.2) give I1 (8.2.5) Y11 = V1 V2 =0 and Y21 =

I2 V1 V2 =0

(8.2.6)

Similarly, with a source connected at port 2 and a short circuit at port 1, I1 (8.2.7) Y12 = V2 V1 =0 and Y22

I2 = V2 V1 =0

(8.2.8)

Equations (8.2.5) through (8.2.8) define the admittance parameters of a twoport network. Example 8.5 Find the admittance parameters of the circuit shown in Figure 8.7. SOLUTION If V2 is zero, I1 is equal to 0.05 V1 and I2 is −0.05 V1 . Hence, from (8.2.5) and (8.2.6), I1 0.05 V1 Y11 = = = 0.05 S V1 V2 =0 V1 and Y21

I2 −0.05 V1 = = = −0.05 S V1 V2 =0 V1

Similarly, with a source connected at port 2 and port 1 having a short circuit, I2 = −I1 = 0.05 V2

0.05 S I1

+ V1 −

+ V2 −

Figure 8.7 Circuit for Example 8.5.

I2

291

ADMITTANCE PARAMETERS

Hence, from (8.2.7) and (8.2.8), Y12

I1 −0.05 V2 = = = −0.05 S V2 V1 =0 V2

Y22

I2 0.05 V2 = = = 0.05 S V2 V1 =0 V2

and

Therefore,

Y11 Y21

Y12 Y22

0.05 −0.05 = −0.05 0.05

Again we find that Y11 is equal to Y22 because this circuit is symmetrical. Similarly, Y12 is equal to Y21 because it is reciprocal. Example 8.6 Find the admittance parameters for the two-port network shown in Figure 8.8. SOLUTION Assuming that a source is connected at port 1 and port 2 has a short circuit, we find that I1 =

0.1(0.2 + 0.025) 0.0225 V1 = V1 0.1 + 0.2 + 0.025 0.325

A

and if voltage across 0.2 S is VN , then VN =

0.0225 I1 V1 = V1 = 0.2 + 0.025 0.225 × 0.325 3.25

V

Therefore, I2 = −0.2 VN = −

0.2 V1 3.25

0.1 S I1

0.2 S

+ V1 −

Figure 8.8

A

+

0.025 S

V2 −

Two-port network for Example 8.6.

I2

292

TWO-PORT NETWORKS

Hence, from (8.2.5) and (8.2.6), I1 0.0225 Y11 = = = 0.0692 S V1 V2 =0 0.325 and Y21 =

I2 0.2 = −0.0615 S =− V1 V2 =0 3.25

Similarly, with a source at port 2 and port 1 having a short circuit, I2 =

0.2(0.1 + 0.025) 0.025 V1 = V2 0.2 + 0.1 + 0.025 0.325

A

and if voltage across 0.1 S is VM , then VM =

I2 2 V2 0.025 = V2 = 0.1 + 0.025 0.125 × 0.325 3.25

Therefore, I1 = −0.1 VM = −

0.2 V2 3.25

V

A

Hence, from (8.2.7) and (8.2.8), I1 0.2 Y12 = =− = −0.0615 S V2 V1 =0 3.25 and Y22 Therefore,

I2 0.025 = = = 0.0769 S V2 V1 =0 0.325

Y11 Y21

Y12 Y22

=

0.0692 −0.0615

−0.0615 0.0769

As expected, Y12 = Y21 but Y11 = Y22 . This is because the given circuit is reciprocal but is not symmetrical. Example 8.7 Find the admittance parameters of the two-port network shown in Figure 8.9. SOLUTION Assuming that a source is connected at port 1 and port 2 has a short circuit, we find that 0.1(0.2 + 0.025) I1 = 0.05 + A V1 = 0.1192 V1 0.1 + 0.2 + 0.025

293

ADMITTANCE PARAMETERS 0.05 S 0.1 S I1

0.2 S +

+ V1

0.025 S

I2

V2 −

−

Figure 8.9

Two-port network for Example 8.7.

and if current through 0.05 S is IN , then IN =

0.05 I1 = 0.05 V1 0.05 + [0.1(0.2 + 0.025)/(0.1 + 0.2 + 0.025)]

A

Current through 0.1 S is I1 − IN = 0.0692 V1 . Using the current-division rule, current IM through 0.2 S is found as follows: IM =

0.2 0.0692 V1 = 0.0615 V1 0.2 + 0.025

A

Hence, I2 = −(IN + IM ) = −0.1115 V1 A. Now, from (8.2.5) and (8.2.6), Y11

I1 = = 0.1192 S V1 V2 =0

and Y21 =

I2 = −0.1115 S V1 V2 =0

Similarly, with a source at port 2 and port 1 having a short circuit, current I2 at port 2 is

0.2(0.1 + 0.025) I2 = 0.05 + V2 = 0.1269 V2 0.2 + 0.1 + 0.025

A

and current IN through 0.05 S can be found as follows: IN =

0.05 I2 = 0.05 V2 0.05 + [0.2(0.1 + 0.025)/(0.2 + 0.1 + 0.025)]

A

294

TWO-PORT NETWORKS 0.05 S + V1 −

I1

+ V2 −

I2

+

+

V1 −

V2

0.1 S

0.2 S

− +

+ V1

V2

0.025 S

−

Figure 8.10

−

Circuit for Example 8.7.

Current through 0.2 S is I2 − IN = 0.0769 V2 . Using the current-division rule one more time, the current IM through 0.1 S is found as follows: IM =

0.1 0.0769 V2 = 0.0615V2 0.1 + 0.025

A

Hence, I1 = −(IN + IM ) = −0.1115 V2 A. Therefore, from (8.2.7) and (8.2.8), Y12

I1 = = −0.1115 S V2 V1 =0

Y22

I2 = = 0.1269 S V2 V1 =0

and

Therefore,

Y11 Y21

Y12 Y22

=

0.1192 −0.1115

−0.1115 0.1269

As expected, Y12 = Y21 but Y11 = Y22 . This is because the given circuit is reciprocal but is not symmetrical. Further, we find that the admittance parameters obtained in Example 8.7 are equal to the sum of the corresponding impedance parameters of Examples 8.5 and 8.6. This is because when the circuits of these two examples are connected in parallel, we end up with the circuit of Example 8.7, shown in Figure 8.10. Example 8.8 Find the admittance parameters of a transmission line of length l, shown in Figure 8.11.

295

ADMITTANCE PARAMETERS l I1

+

+ Z0, γ

V1

−

I2

V2

−

Figure 8.11 Setup for Example 8.8.

SOLUTION This circuit is symmetrical because interchanging port 1 and port 2 does not affect it. Therefore, Y22 must be equal Y11 . Further, if voltage V at port 1 produces a short-circuit current I at port 2, voltage V at port 2 will produce current I at port 1. Hence, it is a reciprocal circuit. Therefore, Y12 will be equal to Y21 . Assume that a source is connected at port 1 when the other port has a short circuit. If Vin is the incident voltage at port 1, it will appear as Vin e−γl at port 2. Since the reflection coefficient of a short circuit is equal to −1, reflected voltage at this port is 180◦ out of phase with incident voltage. Therefore, the reflected voltage reaching at port 1 is −Vin e−2γl . Hence, V1 = Vin − Vin e−2γl V2 = 0 Vin (1 + e−2γl ) I1 = Z0 and I2 = −

2Vin −γl e Z0

Therefore, Y11

e+γl + e−γl 1 I1 (Vin /Z0 )(1 + e−2γl ) = = = = −2γl +γl −γl V1 V2 =0 Vin (1 − e ) Z0 (e − e ) Z0 tanh γl

Y21

I2 −(2Vin /Z0 )e−γl 2 1 = = =− =− −2γl +γl −γl V1 V2 =0 Vin (1 − e ) Z0 (e − e ) Z0 sinh γl

and

For a lossless line, γ = j β and therefore Y11 =

1 j Z0 tan βl

and Y21 = −

1 1 =j j Z0 sin βl Z0 sin βl

296

TWO-PORT NETWORKS

8.3 HYBRID PARAMETERS Reconsider the two-port network of Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 at port 1 can be expressed in terms of current I1 at port 1 and voltage V2 at port 2: V1 = h11 I1 + h12 V2 (8.3.1) Similarly, we can write I2 in terms of I1 and V2 : I2 = h21 I1 + h22 V2

(8.3.2)

Since V1 and V2 are in volts while I1 and I2 are in amperes, parameter h11 must be in ohms, h12 and h21 must be dimensionless, and h22 must be in siemens. Therefore, these are called the hybrid parameters. Using the matrix representation, we can write h11 h12 I1 V1 = (8.3.3) I2 h21 h22 V2 Hybrid parameters are especially important in transistor circuit analysis. These parameters are determined as follows. If port 2 has a short circuit, V2 will be zero. In this condition, (8.3.1) and (8.3.2) give V1 h11 = (8.3.4) I1 V2 =0 and h21 =

I2 I1 V2 =0

Similarly, with a source connected at port 2 while port 1 is open, V1 h12 = V2 I1 =0 and h22

I2 = V2 I1 =0

(8.3.5)

(8.3.6)

(8.3.7)

Thus, parameters h11 and h21 represent the input impedance and the forward current gain, respectively, when a short circuit is at port 2. Similarly, h12 and h22 represent the reverse voltage gain and the output admittance, respectively, when port 1 has an open circuit. Because of this mix, these are called hybrid parameters. In transistor circuit analysis, these are generally denoted by hi , hf , hr , and h0 , respectively.

297

HYBRID PARAMETERS 12 Ω I1

3Ω

+

+

V1

6Ω

I2

V2 −

−

Figure 8.12

Two-port network for Example 8.9..

Example 8.9 Find the hybrid parameters of the two-port network shown in Figure 8.12. SOLUTION With a short circuit at port 2, 6×3 V1 = I1 12 + = 14I1 6+3 and using the current-divider rule, we find that I2 = −

6 2 I1 = − I1 6+3 3

Therefore, from (8.3.4) and (8.3.5), h11

V1 = = 14 I1 V2 =0

and h21 =

I2 2 =− I1 V2 =0 3

Similarly, with a source connected at port 2 while port 1 has an open circuit, V2 = (3 + 6)I2 = 9I2 and V1 = 6I2 because there is no current flowing through a 12- resistor. Hence, from (8.3.6) and (8.3.7), V1 6I2 2 h12 = = = V2 I1 =0 9I2 3 and h22

I2 1 = = S V2 I1 =0 9

298

Thus,

TWO-PORT NETWORKS

h11 h21

h12 h22

=

2 3 1 9

14 − 23

S

8.4 TRANSMISSION PARAMETERS Reconsider the two-port network of Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 and current I1 at port 1 can be expressed in terms of current I2 and voltage V2 at port 2 as follows: V1 = AV2 − BI2

(8.4.1)

Similarly, we can write I1 in terms of I2 and V2 as follows: I1 = CV2 − DI2

(8.4.2)

Since V1 and V2 are in volts while I1 and I2 are in amperes, parameters A and D must be dimensionless, B must be in ohms, and C must be in siemens. Using the matrix representation, (8.4.2) can be written as follows:

V1 I1

A = C

B D

V2 −I2

(8.4.3)

Transmission parameters (also known as elements of chain matrix ) are especially important for analysis of circuits connected in cascade. These parameters are determined as follows. If port 2 has a short circuit, V2 will be zero. Under this condition, (8.4.1) and (8.4.2) give V1 B= (8.4.4) −I2 V2 =0 and

I1 D= −I2 V2 =0

(8.4.5)

Similarly, with a source connected at the port 1 while port 2 is open, we find that V1 A= (8.4.6) V2 I2 =0 and

I1 C= V2 I2 =0

(8.4.7)

299

TRANSMISSION PARAMETERS

I1

1Ω +

+

V1

V2

−

−

Figure 8.13

I2

Network for Example 8.10.

Example 8.10 Determine the transmission parameters of the network shown in Figure 8.13. SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), I2 = −I1 and V1 = I1 V Therefore, from (8.4.4) and (8.4.5), B= and

V1 = 1 −I2 V2 =0

I1 D= =1 −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), V2 = V1 and I1 = 0 Now, from (8.4.6) and (8.4.7),

V1 =1 A= V2 I2 =0

and C=

I1 =0 V2 I2 =0

Hence, the transmission matrix of this network 1 A B = 0 C D

is 1 1

Example 8.11 Determine the transmission parameters of the network shown in Figure 8.14. SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), I2 = −I1 and V1 = 0V

300

TWO-PORT NETWORKS I1

+

+ jω S

V1 −

Figure 8.14

I2

V2 −

Network for Example 8.11.

Therefore, from (8.4.4) and (8.4.5), V1 = 0 B= −I2 V2 =0 and D=

I1 =1 −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), A V2 = V1 and I1 = j ωV1 Now, from (8.4.6) and (8.4.7), V1 =1 A= V2 I2 =0 and

I1 C= = jω V2 I2 =0

Hence, transmission matrix of this network is 1 A B = jω C D

S

0 1

Example 8.12 Determine the transmission parameters of the network shown in Figure 8.15. 1Ω I1

1Ω

+ V1 −

Figure 8.15

+ jω S

V2 −

Network for Example 8.12.

I2

301

TRANSMISSION PARAMETERS

SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), we find that 1 2 + jω V1 = 1 + I1 = I1 1 + jω 1 + jω and 1 1 jω I2 = − I1 = − I1 1/j ω + 1 1 + jω Therefore, from (8.4.4) and (8.4.5), V1 = 2 + jω B= −I2 V2 =0 and

I1 D= = 1 + jω −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), 1 1 + jω I1 = I1 V1 = 1 + jω jω and V2 =

1 I1 jω

Now, from (8.4.6) and (8.4.7), V1 = 1 + jω A= V2 I2 =0 and C= Hence,

A C

B D

I1 = jω V2 I2 =0

1 + jω = jω

S

2 + jω 1 + jω

Example 8.13 Find the transmission parameters of the transmission line shown in Fig. 8.16. SOLUTION Assume that a source is connected at port 1 while the other port has a short circuit. If Vin is incident voltage at port 1, it will be Vin e−γl at port 2.

302

TWO-PORT NETWORKS l I1

+

+ Z0, γ

V1

−

Figure 8.16

I2

V2

−

Transmission line for Example 8.13.

Since the reflection coefficient of the short circuit is −1, the reflected voltage at this port is 180◦ out of phase with the incident voltage. Therefore, the reflected voltage reaching at port 1 is −Vin e−2γl . Hence, V1 = Vin − Vin e−2γl V2 = 0 Vin (1 + e−2γl ) I1 = Z0 and I2 = −

2Vin −γl e Z0

Therefore, from (8.4.4) and (8.4.5), V1 Z0 eγl − e−γl −2λl B= = (1 − e ) = Z 0 −I2 V2 =0 2e−γl 2

= Z0 sinh γl

and

I1 1 + e−2λl eγl + e−γl D= = = = cosh γl −I2 V2 =0 2e−λl 2

Now assume that port 2 has an open circuit while the source is still connected at port 1. If Vin is incident voltage at port 1, Vin e−γl is at port 2. Since the reflection coefficient of an open circuit is +1, the reflected voltage at this port is equal to the incident voltage. Therefore, the reflected voltage reaching at port 1 is Vin e−2γl . Hence, V1 = Vin + Vin e−2γl V2 = 2Vin e−γl Vin I1 = (1 − e−2γl ) Z0 and I2 = 0

303

TRANSMISSION PARAMETERS

Now, from (8.4.6) and (8.4.7), 1 + e−2γl V1 = = cosh γl A= V2 I2 =0 2e−γl and

I1 1 − e−2γl 1 C= = = sinh γl −γl V2 I2 =0 2Z0 e Z0

Hence, the transmission matrix of a finite-length transmission line is cosh γl Z0 sinh γl A B = 1 C D sinh γl cosh γl Z0 For a lossless line, γ = j β, and therefore it simplifies to cos βl j Z0 sin βl A B = 1 C D j sin βl cos βl Z0 An analysis of results obtained in Examples 8.10 to 8.13 indicates that the following condition holds for all four circuits: AD − BC = 1

(8.4.8)

This is because these circuits are reciprocal. In other words, if a given circuit is known to be reciprocal, (8.4.8) must be satisfied. Further, we find that transmission parameter A is equal to D in all four cases. This always happens when a given circuit is reciprocal. In Example 8.11, A and D are real, B is zero, and C is imaginary. For a lossless line in Example 8.13, A and D simplify to real numbers while C and D become purely imaginary. This characteristic of the transmission parameters is associated with any lossless circuit. A comparison of the circuits in Examples 8.10 to 8.12 reveals that the two-port network of Example 8.12 can be obtained by cascading that of Example 8.10 on the two sides of Example 8.11, as shown in Figure 8.17. 1Ω I1

1Ω +

+

V1 −

V2 −

jω S 8.10

Figure 8.17

8.11

8.10

Two-port network for Example 8.12.

I2

304

TWO-PORT NETWORKS

Therefore, the chain (or transmission) matrix for the network shown in Example 8.12 can be obtained after multiplying three chain matrices as follows: 1 1 1 0 1 1 1 1 1 1 · · = · 0 1 jω 1 0 1 0 1 jω jω + 1 1 + jω 2 + jω = jω 1 + jω This shows that chain matrices are convenient in the analysis and design of networks connected in cascade. 8.5 CONVERSION OF IMPEDANCE, ADMITTANCE, CHAIN, AND HYBRID PARAMETERS One type of network parameter can be converted into another via the respective defining equations. For example, the admittance parameters of a network can be found from its impedance parameters as follows. From (8.2.3) and (8.1.3), we find that −1 I1 Y11 V1 Z11 V1 Y12 Z12 = = I2 Y21 Y22 V2 Z21 Z22 V2 Hence, Y11 Y21

Y12 Y22

=

Z11 Z21

Z12 Z22

−1

=

1 Z11 Z22 − Z12 Z21

Z22 −Z21

−Z12 Z11

Similarly, (8.3.3) can be rearranged as follows: D AD − BC − I1 V1 B = B I2 V2 1 A − B B Hence, D AD − BC − Y11 Y12 B = B Y21 Y22 1 A − B B Relations between other parameters can be found following a similar procedure. These relations are given in Table 8.1.

8.6 SCATTERING PARAMETERS As illustrated in preceding sections, Z-parameters are useful in analyzing series circuits while Y -parameters simplify the analysis of parallel (shunt) connected circuits. Similarly, transmission parameters are useful for chain or cascade circuits.

305

SCATTERING PARAMETERS

TABLE 8.1 Conversions Among the Impedance, Admittance, Chain, and Hybrid Parameters Y22 Y11 Y22 − Y12 Y21 −Y12 = Y11 Y22 − Y12 Y21 −Y21 = Y11 Y22 − Y12 Y21 Y11 = Y11 Y22 − Y12 Y21

A C AD − BC = C 1 = C D = C

h11 h22 − h12 h21 h22 h12 = h22 −h21 = h22 1 = h22

Z11 =

Z11 =

Z11 =

Z12

Z12

Z12

Z21 Z22

Z22 Z11 Z22 − Z12 Z21 −Z12 = Z11 Z22 − Z12 Z21 −Z21 = Z11 Z22 − Z12 Z21 Z11 = Z11 Z22 − Z12 Z21

Z21 Z22

D B −(AD − BC) = B −1 = B A = B

Z21 Z22

1 h11 −h12 = h11 h21 = h11 h11 h22 − h12 h21 = h11

Y11 =

Y11 =

Y11 =

Y12

Y12

Y12

Y21 Y22

Z11 Z21 Z11 Z22 − Z12 Z21 B= Z21 1 C= Z21 Z22 D= Z21 A=

Z11 Z22 − Z12 Z21 Z22 Z12 = Z22 −Z21 = Z22 1 = Z22

Y21 Y22

−Y22 Y21 −1 B= Y21 −(Y11 Y22 − Y12 Y21 ) C= Y21 −Y11 D= Y21 A=

1 Y11 −Y12 = Y11 Y21 = Y11 Y11 Y22 − Y12 Y21 = Y11

h11 =

h11 =

h12

h12

h21 h22

h21 h22

Y21 Y22

−(h11 h22 − h12 h21 ) h21 −h11 B= h21 −h22 C= h21 −1 D= h21 A=

B D AD − BC = D −1 = D C = D

h11 = h12 h21 h22

306

TWO-PORT NETWORKS

However, the characterization procedure of these parameters requires an open or short circuit at the other port. This extreme reflection makes it very difficult (and in certain cases, impossible) to determine the parameters of a network at radio and microwave frequencies. Therefore, a new representation based on traveling waves is defined. This is known as the scattering matrix of the network. Elements of this matrix are known as scattering parameters. Figure 8.18 shows a network along with incident and reflected waves at its two ports. We adopt a convention of representing the incident wave by ai and the reflected wave by bi at the ith port. Hence, a1 is an incident wave, while b1 is a reflected wave at port 1. Similarly, a2 and b2 represent incident and reflected waves at port 2, respectively. Assume that a source is connected at port 1 that produces the incident wave a1 . A part of this wave is reflected back at the input (due to impedance mismatch), while the remaining signal is transmitted through the network. It may change in magnitude as well as in phase before emerging at port 2. Depending on the termination at this port, part of the signal is reflected back as input to port 2. Hence, reflected wave b1 depends on incident signals a1 and a2 at the two ports. Similarly, emerging wave b2 also depends on a1 and a2 . Mathematically, b1 = S11 a1 + S12 a2

(8.6.1)

b2 = S21 a1 + S22 a2

(8.6.2)

Using the matrix notation, we can write b1 S11 = b2 S21 or

S12 S22

a1 a2

(8.6.3)

[b] = [S][a]

(8.6.4)

where [S] is called the scattering matrix of the two-port network; Sij are known as the scattering parameters of this network, and ai represents the incident wave at the ith port and bi represents the reflected wave at the ith port. If port 2 is matched terminated while a1 is incident at port 1, a2 is zero. In this condition, (8.6.1) and (8.6.2) give b1 S11 = (8.6.5) a1 a2 =0

a1 Port 1 b 1

Two-port network

b2

Port 2

a2

Figure 8.18 Two-port network with associated incident and reflected waves.

SCATTERING PARAMETERS

and S21

b2 = a1 a2 =0

307

(8.6.6)

Similarly, with a source connected at port 2 while port 1 is terminated by a matched load, we find that b1 S12 = (8.6.7) a2 a1 =0 and S22

b2 = a2 a1 =0

(8.6.8)

Hence, Sii is the reflection coefficient i at the ith port when the other port is matched terminated. Sij is the forward transmission coefficient of the j th port if i is greater than j , whereas it represents the reverse transmission coefficient if i is less than j with the other port terminated by a matched load. We have not yet defined ai and bi in terms of voltage, current, or power. To that end, we write steady-state total voltage and current at the ith port as follows: Vi = Viin + Viref

(8.6.9)

and Ii =

1 (V in − Viref ) Z0i i

(8.6.10)

where superscripts “in” and “ref” represent the incident and reflected voltages, respectively. Z0i is the characteristic impedance at the ith port. Equations (8.6.9) and (8.6.10) can be solved to find incident and reflected voltages in terms of total voltage and current at the ith port. Hence, Viin = 12 (Vi + Z0i Ii )

(8.6.11)

Viref = 12 (Vi − Z0i Ii )

(8.6.12)

and

Assuming both of the ports to be lossless so that Z0i is a real quantity, the average power incident at the ith port is in ∗ 1 1 in in ∗ 1 in in Vi = |V in |2 (8.6.13) Pi = Re Vi (Ii ) = Re Vi 2 2 Z0i 2Z0i i and average power reflected from the ith port is ref ∗ 1 1 ref ref ∗ 1 ref ref Vi = |V ref |2 Pi = Re Vi (Ii ) = Re Vi 2 2 Z0i 2Z0i i

(8.6.14)

308

TWO-PORT NETWORKS

The ai and bi are defined in such a way that the squares of their magnitudes represent the power flowing in respective directions. Hence, Viin 1 Vi + Z0i I i 1 Vi = + Z0i Ii = √ (8.6.15) ai = √ √ √ 2 2Z 0i 2Z0i Z0i 2 2 and V ref 1 = bi = √ i 2 2Z 0i

Vi − Z0i I i √ 2Z0i

1 = √ 2 2

Vi − Z0i Ii √ Z0i

(8.6.16)

Therefore, units of ai and bi are √

√ volt watt = √ = ampere · ohm ohm

Power available from the source, Pavs , at port 1 is Pavs = |a1 |2 power reflected from port 1, Pref , is Pref = |b1 |2 and power delivered to the port (and hence to the network), Pd , is Pd = Pavs − Pref = |a1 |2 − |b1 |2 Consider the circuit arrangement shown in Figure 8.19. There is a voltage source VS1 connected at port 1 while port 2 is terminated by load impedance ZL . The source impedance is ZS . Various voltages, currents, and waves are as depicted at the two ports of this network. Further, it is assumed that the characteristic impedances at port 1 and port 2 are Z01 and Z02 , respectively. Input impedance Z1 at port 1 of the network is defined as the impedance across its terminals when port 2 is terminated by load ZL while source VS1 along with ZS are disconnected. Similarly, output impedance Z2 at port 2 of the network is defined as the impedance across its terminals with load ZL disconnected and voltage source VS1 replaced by a short circuit. Hence, source impedance ZS terminates ZS VS1

I1 a1 Port 1

b1

+ V1 −

Figure 8.19 terminated.

+ Two-port network

V2 −

I2 b2

Port 2

a2 ZL

Two-port network with a voltage source connected at port 1 and port 2 is

SCATTERING PARAMETERS

309

port 1 of the network in this case. Input impedance Z1 and output impedance Z2 are responsible for input reflection coefficient 1 and output reflection coefficient 2 , respectively. Hence, the ratio of b1 to a1 represents 1 while that of b2 to a2 is 2 . For the two-port network, we can write b1 = S11 a1 + S12 a2

(8.6.17)

b2 = S21 a1 + S22 a2

(8.6.18)

and

The load reflection coefficient L is L =

ZL − Z 02 a2 = ZL + Z02 b2

(8.6.19)

Note that b2 leaves port 2, and therefore it is incident on the load ZL . Similarly, the wave reflected back from the load enters port 2 as a2 . The source reflection coefficient S is found as S =

Z S − Z01 a1 = ZS + Z01 b1

(8.6.20)

Since b1 leaves port 1 of the network, it is the incident wave on ZS , while a1 is the reflected wave. The input and output reflection coefficients are 1 =

Z1 − Z01 b1 = Z1 + Z01 a1

(8.6.21)

2 =

Z2 − Z02 b2 = Z2 + Z02 a2

(8.6.22)

and

Dividing (8.6.17) by a1 and then using (8.6.21), we find that Z1 − Z01 a2 b1 = 1 = = S11 + S12 a1 Z1 + Z01 a1

(8.6.23)

Now, dividing (8.6.18) by a2 and then combining with (8.6.19), we get 1 a1 1 − S22 L a1 b2 = S22 + S21 = ⇒ = a2 a2 L a2 S21 L

(8.6.24)

From (8.6.23) and (8.6.24), 1 = S11 +

S12 S21 L 1 − S22 L

(8.6.25)

310

TWO-PORT NETWORKS

If a matched load is terminating port 2, then L = 0 and (7.6.25) simplifies to 1 = S11 . Similarly, from (8.6.18) and (8.6.22), Z2 − Z02 a1 b2 = 2 = = S22 + S21 a2 Z2 + Z02 a2

(8.6.26)

From (8.6.17) and (8.6.20), we have 1 a2 1 − S11 S a2 b1 = S11 + S12 = ⇒ = a1 a1 S a1 S12 S

(8.6.27)

Substituting (8.6.27) into (8.6.26), we get 2 = S22 +

S21 S12 S 1 − S11 S

(8.6.28)

If ZS is equal to Z01 , port 1 is matched and S = 0. Therefore, (8.6.28) simplifies to 2 = S22 Hence, S11 and S22 can be found by evaluating the reflection coefficients at respective ports while the other port is matched terminated. Let us determine the other two parameters, S21 and S12 , of the two-port network. Starting with (8.6.6) for S21 , we have S21

b2 = a1 a2 =0

(8.6.6)

Now, a2 is found from (8.6.15) with i as 2 and forcing it to zero, we get a2 =

1 2

V2 + Z02 I2 √ 2Z02

= 0 ⇒ V2 = −Z02 I2

(8.6.29)

Substituting (8.6.29) in the expression for b2 that is obtained from (8.6.16) with i as 2, we find that 1 b2 = 2

V2 − Z02 I2 √ 2Z02

= −I2

Z02 2

(8.6.30)

An expression for a1 is obtained from (8.6.15) with i = 1. It simplifies for ZS equal to Z01 as follows: 1 a1 = 2

V1 + Z01 I1 √ 2Z01

VS1 = √ 2 2Z01

(8.6.31)

311

SCATTERING PARAMETERS

S21 is obtained by substituting (8.6.30) and (8.6.31) into (8.6.6) as follows: √ √ b2 − Z02 I2 / 2 2V2 Z01 S21 = = = (8.6.32) √ a1 a2 =0 VS1 Z02 VS1 /2 2Z01 Following a similar procedure, S12 may be found as b1 2V1 Z02 = S12 = a2 a1 =0 VS2 Z01

(8.6.33)

An analysis of S-parameters indicates that |b1 |2 Pavs − Pd 2 = |S11 | = 2 |a1 | a2 =0 Pavs

(8.6.34)

where Pavs is power available from the source and Pd is power delivered to port 1. These two powers will be equal if the source impedance is conjugate of Z1 ; that is, the source is matched with port 1. Similarly, from (8.6.32), √ √ PAVN Z02 (I2 / 2)2 Z02 (I2 / 2)2 2 = = 1 (8.6.35) |S21 | = 1 √ √ 2 2 Pavs ( 4 Z01 )(VS1 / 2) [(1/2Z01 )(VS1 / 2) ] 2 where PAVN is power available at port 2 of the network. It will be equal to power delivered to a load that is matched to the port. This power ratio of (8.6.35) may be called the transducer power gain. Following a similar procedure, it may be found that |S22 |2 represents the ratio of power reflected from port 2 to power available from the source at port 2, while port 1 is terminated by a matched load ZS and |S12 |2 represents a reverse transducer power gain. Shifting the Reference Planes Consider the two-port network shown in Figure 8.20. Assume that ai and bi are incident and reflected waves, respectively, at unprimed reference planes of the ith A

A′

B

x1

x2 a1 Port 1

b2

Two-port network b1

Port 2

a2

l1 A′

B′

l2 A

B

Figure 8.20 Two-port network with two reference planes on each side.

B′

312

TWO-PORT NETWORKS

port. We use unprimed S-parameters for this case. Next, consider that plane A–A is shifted by a distance l1 to A –A . At this plane, a1 and b1 represent inward- and outward-traveling waves, respectively. Similarly, a2 and b2 represent inward- and outward-traveling waves, respectively, at plane B –B . We denote the scattering parameters at primed planes by a prime on each as well. Hence, b1 S11 S12 a1 = (8.6.36) b2 S21 S22 a2 and

b1 b2

=

S11 S21

S12 S22

a1 a2

(8.6.37)

Wave b1 is delayed in phase by βl1 as it travels from A to A . This means that b1 is ahead in phase with respect to b1 . Hence, b1 = b1 ej βl1

(8.6.38)

Wave a1 comes from A to plane A. Therefore, it has a phase delay of βl1 with respect to a1 . Mathematically, a1 = a1 e−j βl1

(8.6.39)

On the basis of similar considerations at port 2, we can write b2 = b2 ej βl2

(8.6.40)

a2 = a2 e−j βl2

(8.6.41)

and

Substituting for b1 , a1 , b2 , and a1 from (8.6.38) to (8.6.41) into (8.6.36), we get

b1 ej βl1 b2 ej βl2

S11 = S21

S12 S22

a1 e−j βl1

(8.6.42)

a2 e−j βl2

We can rearrange this equation as follows:

b1 b2

=

S11 e−j 2βl1

S12 e−j β(l1 +l2 )

S21 e−j β(l1 +l2 )

S22 e−j 2βl2

a1 a2

(8.6.43)

Now, on comparing (8.6.43) with (8.6.37), we find that

S11 S21

S12 S22

S11 e−j 2βl1 = S21 e−j β(l1 +l2 )

S12 e−j β(l1 +l2 ) S22 e−j 2βl2

(8.6.44)

313

SCATTERING PARAMETERS I1

I2

Two-port network

V1

Figure 8.21

V2

Network for Example 8.14.

Following a similar procedure, one can find the other relation as follows: j 2βl1 j β(l1 +l2 ) S12 e S11 e S11 S12 (8.6.45) = j β(l1 +l2 ) j 2βl2 S21 S22 S21 e S22 e Example 8.14 The total voltages and currents at two ports of a network are found as follows: V1 = 10 0◦ V, I1 = 0.1 40◦ A, V2 = 12 30◦ V, and I2 = 0.15 100◦ A. Determine the incident and reflected voltages, assuming that the characteristic impedance is 50 at both its ports (see Figure 8.21). SOLUTION From (8.6.11) and (8.6.12) with i = 1, we find that ◦

◦

◦

◦

V1in = 12 (10 0 + 50 × 0.1 40 ) = 6.915 + j 1.607 V and V1ref = 12 (10 0 − 50 × 0.1 40 ) = 3.085 − j 1.607 V Similarly, with i = 2, incident and reflected voltages at port 2 are found to be ◦

◦

◦

◦

V2in = 12 (12 30 + 50 × 0.15 100 ) = 4.545 + j 6.695 V and V2ref = 12 (12 30 − 50 × 0.15 100 ) = 5.845 − j 0.691 V Example 8.15 Find the S-parameters of a series impedance Z connected between the two ports, as shown in Figure 8.22.

Z Z0

Z0

Figure 8.22

Setup for Example 8.15.

314

TWO-PORT NETWORKS

SOLUTION From (8.6.25), with L = 0 (i.e., port 2 is terminated by a matched load), we find that S11 = 1 |a2 =0 =

(Z + Z0 ) − Z0 Z = (Z + Z0 ) + Z0 Z + 2Z0

Similarly, from (8.6.28), with S = 0 (i.e., port 1 is terminated by a matched load), (Z + Z0 ) − Z0 Z = S22 = 2 |a1 =0 = (Z + Z0 ) + Z0 Z + 2Z0 S21 and S12 are determined from (8.6.32) and (8.6.33), respectively. For evaluating S21 , we connect a voltage source VS1 at port 1 while port 2 is terminated by Z0 , as shown in Figure 8.23. The source impedance is Z0 . Using the voltage-divider formula, we can write V2 =

Z0 VS1 Z + 2Z0

Hence, from (8.6.32), S21 =

2V2 2Z0 2Z0 + Z − Z Z = = =1− VS1 Z + 2Z0 Z + 2Z0 Z + 2Z0

For evaluating S12 , we connect a voltage source VS2 at port 2 while port 1 is terminated by Z0 , as shown in Figure 8.24. The source impedance is Z0 . Using the voltage-divider rule again, we find that V1 =

Z0 VS2 Z + 2Z0

Z0

Z

VS1 ~

V2

V1

Figure 8.23

Setup to determine S21 for Example 8.15.

Z Z0

V1

Figure 8.24

Z0

Z0 V2

~ VS 2

Setup to determine S12 for Example 8.15.

315

SCATTERING PARAMETERS

Now, from (8.6.33), S12 =

2V1 2Z0 2Z0 + Z − Z Z = = =1− VS2 Z + 2Z0 Z + 2Z0 Z + 2Z0

Therefore,

S11 S21

=

1 1 − 1

1 =

Z Z + 2Z0

S12 S22

where

1 − 1 1

An analysis of the S-parameters indicates that S11 is equal to S22 . This is because the given two-port network is symmetrical. Further, its S12 is equal to S21 as well. This happens because this network is reciprocal. Consider a special case where the series impedance Z is a purely reactive element, which means that Z is equal to j X. In that case, 1 can be written as 1 =

jX j X + 2Z0

Therefore, 2 2 jX + 1 − j X + 2Z 0 0

jX |S11 | + |S21 | = j X + 2Z 2

2

=

X2 (2Z0 )2 + =1 X2 + (2Z0 )2 X2 + (2Z0 )2

Similarly, |S12 |2 + |S22 |2 = 1 On the other hand, ∗ ∗ + S21 S22 = S11 S12

jX 2Z0 2Z0 −j X × + × =0 j X + 2Z0 −j X + 2Z0 j X + 2Z0 −j X + 2Z0

and ∗ ∗ S12 S11 + S22 S21 =

2Z0 −j X jX 2Z0 × + × =0 j X + 2Z0 −j X + 2Z0 j X + 2Z0 −j X + 2Z0

These characteristics of the scattering matrix can be summarized as follows:

∗ Sij Sik

= δj k =

1 0

if j = k otherwise

316

TWO-PORT NETWORKS

Z0

Z0

Y

Figure 8.25

Network for Example 8.16.

When the elements of a matrix satisfy this condition, it is called a unitary matrix. Hence, the scattering matrix of a reactance circuit is unitary. Example 8.16 Find the S-parameters of a shunt admittance Y connected between two ports, as shown in Figure 8.25. SOLUTION From (8.6.25), S11 = 1 |L =0 = =

Z1 − Z01 1/Y1 − 1/Y0 = Z1 + Z01 1/Y1 + 1/Y0 Y0 − Y1 Y0 − (Y + Y0 ) −Y = = Y0 + Y1 Y0 + (Y + Y0 ) 2Y0 + Y

Similarly, from (8.6.28), S22 = 2 |S =0 = =

Z2 − Z02 1/Y2 − 1/Y0 = Z2 + Z02 1/Y2 + 1/Y0 Y0 − Y2 Y0 − (Y + Y0 ) −Y = = Y0 + Y2 Y0 + (Y + Y0 ) 2Y0 + Y

For evaluating S21 , voltage source VS1 is connected at port 1, and port 2 is terminated by a matched load, as shown in Figure 8.26. Using the voltage-divider rule, V2 =

1/(Y + Y0 ) 1 Y0 VS1 = VS1 = VS1 Z0 + 1/(Y + Y0 ) Z0 (Y + Y0 ) + 1 Y + 2Y0

Now, from (8.6.32), S21

2V2 2Y0 Y = = =1− = 1 + 1 VS1 L =0 Y + 2Y0 Y + 2Y0

Z0 VS1 ~

Figure 8.26

V1

Y

V2

Z0

Setup to determine S21 for Example 8.16.

317

SCATTERING PARAMETERS

Similarly, when VS2 in series with Z0 is connected at port 2 while port 1 is match-terminated, S12 can be found from (8.6.33) as follows: 2V1 2Y0 Y = =1− = 1 + 1 S12 = VS2 S =0 Y + 2Y0 Y + 2Y0 where 1 = −

Y Y + 2Y0

As expected, S11 = S22 and S12 = S21 , because this circuit is symmetrical as well as reciprocal. Further, for Y = j B, the network becomes lossless. In that case, |S11 |2 + |S21 |2 =

B2 (2Y0 )2 + =1 B 2 + (2Y0 )2 B 2 + (2Y0 )2

and ∗ ∗ + S21 S22 = S11 S12

−jB 2Y0 2Y0 jB × + × =0 jB + 2Y0 −jB + 2Y0 jB + 2Y0 −jB + 2Y0

Hence, with Y as jB, the scattering matrix is unitary. In fact, it can easily be proved that the scattering matrix of any lossless two-port network is unitary. Example 8.17 An ideal transformer is designed to operate at 500 MHz. It has 1000 turns on its primary side and 100 turns on its secondary side. Assuming that it has a 50- connector on each side, determine its S-parameters (see Figure 8.27). SOLUTION For an ideal transformer, V1 −I2 = =n V2 I1

and

V1 (−I2 ) V1 /I1 Z1 = = n2 = V2 I 1 V2 /(−I2 ) Z2

Now, following the procedure used in preceding examples, we find that b1 Z0 n2 − Z0 n2 − 1 99 S11 = = | = = = 1 a2 =0 2 2 a1 a2 =0 Z0 n + Z0 n +1 101

I1 Port 1

n :1 + + V2 −

V1 − 1000 : 100

Figure 8.27 Transformer for Example 8.17.

I2 Port 2

318

TWO-PORT NETWORKS

and voltage V1 on the primary side of the transformer due to a voltage source VS1 with its internal impedance Z0 can be determined by the voltage-divider rule as follows: V1 =

Z0 n2 VS1 Z0 n2 + Z0

Hence, S21

V1 n2 nV2 = 2 = VS1 n +1 VS1

⇒

2V2 2n 20 = = 2 = VS1 a2 =0 n +1 101

Similarly, S22 and S12 are determined after connecting a voltage source VS2 with its internal impedance Z0 at port 2. Port 1 is terminated by a matched load this time. Therefore, b2 (Z0 /n2 ) − Z0 1 − n2 99 = 2 |a1 =0 = = =− S22 = 2 2 a2 a1 =0 (Z0 /n ) + Z0 1+n 101 and V2 =

Z0 /n2 V2 1/n2 V1 /n 1 VS2 ⇒ = = = 2 2 2 (Z0 /n ) + Z0 VS2 (1/n ) + 1 1+n VS2

Hence, S12 Thus,

2V1 2n 20 = = 2 = VS2 a1 =0 n +1 101 S11 S21

S12 S22

=

99 101 20 101

20 101 99 − 101

This time, S11 is different from S22 because the network is not symmetrical. However, S12 is equal to S21 because it is reciprocal. Further, an ideal transformer is lossless. Hence, its scattering matrix must be unitary. Let us verify if that is the case. 10,201 99 2 20 2 2 2 + = |S11 | + |S21 | = =1 101 101 10,201 20 2 99 2 10,201 + − = =1 |S12 |2 + |S22 |2 = 101 101 10,201 20 20 99 99 ∗ ∗ × + × − S11 S12 + S21 S22 = =0 101 101 101 101 and ∗ S12 S11

+

∗ S22 S21

99 99 20 20 × + − =0 = × 101 101 101 101

Hence, the scattering matrix is indeed unitary.

319

SCATTERING PARAMETERS l I1 a1

I2 +

+

b2

Z0, γ V1

V2

−

−

b1

a2

Figure 8.28 Transmission line network for Example 8.18.

Example 8.18 Find the scattering parameters of the transmission line network shown in Figure 8.28. SOLUTION With port 2 match-terminated, wave a1 entering port 1 emerges as a1 e−γl at port 2. This wave is absorbed by the termination, making a2 zero. Further, b1 is zero as well because Z1 is equal to Z0 in this case. Hence, b2 = a1 e−γl Similarly, when a source is connected at port 2 while port 1 is match-terminated, we find that a1 and b2 are zero and b1 = a2 e−γl Hence,

S11 S21

S12 S22

=

0 e−γl

e−γl 0

As expected, S11 = S22 and S12 = S21 , because this circuit is symmetrical as well as reciprocal. Further, for γ = j β, the network becomes lossless. In that case, 0 e−j βl S11 S12 = −j βl S21 S22 e 0 Obviously, it is a unitary matrix now. Example 8.19 Find the scattering parameters of the two-port network shown in Figure 8.29. SOLUTION With port 2 matched-terminated by a 50- load, impedance Z1 at port 1 is Z1 = j 50 +

50(−j 25) = j 50 + 10 − j 20 = 10 + j 30 50 − j 25

320

TWO-PORT NETWORKS

j50 Ω

Port 1

Port 2

Z01 = 50 Ω

Z02 = 50 Ω

−j 25 Ω

Figure 8.29

Two-port network for Example 8.19.

Therefore, S11 = 1 |a2 =0 =

Z1 − Z01 10 + j 30 − 50 ◦ = = 0.74536 116.565 Z1 + Z01 10 + j 30 + 50

Similarly, output impedance Z2 at port 2 when port 1 is match-terminated by a 50- load is (50 + j 50)(−j 25) Z2 = = 10 − j 30 50 + j 50 − j 25 and from (8.6.32), S22 = 2 |a1 =0 =

Z2 − Z02 10 − j 30 − 50 = = 0.74536 Z1 + Z02 10 − j 30 + 50

◦

− 116.565

Now, let us connect a voltage source VS1 with its impedance at 50 at port 1 while port 2 is matched, as shown in Figure 8.30. Using the voltage-divider rule, we find that 10 − j 20 V2 = VS1 50 + j 50 + 10 − j 20 Therefore, S21

2V2 10 − j 20 = =2 = 0.66666 VS1 a2 =0 60 + j 30

50 Ω

◦

− 90

V2 j50 Ω

VS1 ~

Z01 = 50 Ω

Port 2 −j25 Ω

Z02 = 50 Ω

Port 1

Figure 8.30

Setup to determine S21 for Example 8.19.

50 Ω

321

SCATTERING PARAMETERS

V1

50 Ω

V2 j50 Ω

Z01 = 50 Ω

Z02 = 50

−j 25 Ω

~

Port 2

Port 1

VS2

Figure 8.31 Setup to determine S12 for Example 8.19.

Now connect the voltage source at port 2 while port 1 is match-terminated as shown in Figure 8.31. Using the voltage-divider rule again, we find that V2 =

10 − j 30 VS2 = 0.4714 50 + 10 − j 30

◦

− 45 V

Using it one more time, V1 is found as follows: V1 =

50 1 V2 = 0.4714 50 + j 50 1 + j1

◦

− 45 = 0.33333

◦

− 90 V

and now from (8.6.33), we get S12

2V1 = = 2 × 0.33333 VS2 a1 =0

◦

− 90 = 0.66666

◦

− 90

Hence,

S11 S21

S12 S22

0.74536 116.565◦ = 0.66666 − 90◦

0.66666 0.74536

− 90◦ − 116.565◦

In this case, S11 is different from S22 because the network is not symmetrical. However, S12 is equal to S21 because the network is reciprocal. Further, this network is lossless. Hence, its scattering matrix must be unitary. It can be verified as follows: |S11 |2 + |S21 |2 = (0.74536)2 + (0.66666)2 = 0.99999 = 1 |S12 |2 + |S22 |2 = (0.66666)2 + (0.74536)2 = 0.99999 = 1 ◦

◦

∗ ∗ S11 S12 + S21 S22 = (0.74536 116.565 ) × (0.66666 90 )

+ (0.66666

◦

◦

− 90 ) × (0.74536 116.565 ) = 0

322

TWO-PORT NETWORKS

and ∗ ∗ S12 S11 + S22 S21 = (0.66666

+ (0.74536

◦

− 90 ) × (0.74536

◦

− 116.565 )

◦

◦

− 116.565 ) × (0.66666 90 )

=0 Hence, this scattering matrix is indeed unitary. Table 8.2 summarizes the characteristics of parameter matrices of reciprocal and symmetrical two-port networks. The properties of scattering parameters of lossless junctions are listed in Table 8.3. TABLE 8.2 Properties of Parameters of Reciprocal and Symmetrical Two-Port Networks Parameter Matrix Z11 Z12 Z21 Z22 Y11 Y12 Y21 Y22 A B C D S11 S12 S21 S22

TABLE 8.3

Properties Z12 = Z21 Z11 = Z22 Y12 = Y21 Y11 = Y22 AD − BC = 1 A=D S12 = S21 S11 = S22

Properties of Scattering Matrix of Lossless Junctions

Properties

Explanation

Matrix [S] is symmetrical. [S]t = [S], where [S]t is the transpose matrix of [S]. Consequently, Sij = Sj i Matrix [S] is unitary.

[S]a = [S ∗ ]t = [S]−1 , where [S]a is the adjoint matrix of [S], [S ∗ ]t is the conjugate of [S]t , and [S]−1 is the inverse matrix of [S]. Consequently, N 1 for j = k ∗ Sij Sik = δj k = 0 otherwise i=1 Therefore, N i=1

Sij Sij∗ =

N i=1

|Sij |2 = 1,

j = 1, 2, 3, . . . , N

323

CONVERSION TO AND FROM SCATTERING PARAMETERS

8.7 CONVERSION FROM IMPEDANCE, ADMITTANCE, CHAIN, AND HYBRID PARAMETERS TO SCATTERING PARAMETERS, OR VICE VERSA From (8.1.3) we can write 2 2 Vi 1 Zik =√ Zik Ik = √ Z0i 2Z0i 2Z0i k=1 k=1

=

2

Z ik

k=1

Z0i Ik 2

Z0i Ik 2

where i = 1, 2

Therefore, from (8.6.15) and (8.6.16), we have

and

2 1 Z0i ai = (Z ik + δik ) Ik 2 k=1 2

(8.7.1)

2 1 Z0i Ik (Z ik − δik ) bi = 2 k=1 2

(8.7.2)

where δik =

1 for i = k 0 otherwise

Equations (8.7.1) and (8.7.2) can be written in matrix form as follows: [a] = 12 {[Z] + [U ]}[I ]

(8.7.3)

[b] = 12 {[Z] − [U ]}[I ]

(8.7.4)

where Ik =

Z0i Ik 2

and [U ] is the unit matrix. From (8.7.3) and (8.7.4), we have

Hence,

[b] = {[Z] − [U ]}{[Z] + [U ]}−1 [a]

(8.7.5)

[S] = {[Z] − [U ]}{[Z] + [U ]}−1

(8.7.6)

324

TWO-PORT NETWORKS

TABLE 8.4 Conversion from (to) Impedance, Admittance, Chain, or Hybrid to (from) Scattering Parameters ∗ )(Z22 + Z02 ) − Z12 Z21 (Z11 − Z01 (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21 √ 2Z12 R01 R02 = (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21 √ 2Z21 R01 R02 = (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21

∗ + S11 Z01 )(1 − S22 ) + S12 S21 Z01 (Z01 (1 − S11 )(1 − S22 ) − S12 S21 √ 2S12 R01 R02 = (1 − S11 )(1 − S22 ) − S12 S21 √ 2S21 R01 R02 = (1 − S11 )(1 − S22 ) − S12 S21

S11 =

Z11 =

S12

Z12

S21

S22 =

∗ Z02 )

(Z11 + Z01 )(Z22 − − Z12 Z21 (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21

∗ ∗ (1 − Y11 Z01 )(1 + Y22 Z02 ) + Y12 Y21 Z01 Z02 (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02 √ −2Y12 R01 R02 = (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02 √ −2Y21 R01 R02 = (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02

Z21

Z22 =

∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 ∗ + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ −2S12 R01 R02 = ∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ −2S21 R01 R02 = ∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02

S11 =

Y11 =

S12

Y12

S21

S22 =

∗ Y22 Z02 )

∗ Y12 Y21 Z01 Z02

(1 + Y11 Z01 )(1 − + (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02

Y21

∗ (Z02 + S22 Z02 )(1 − S11 ) + S12 S21 Z02 (1 − S11 )(1 − S22 ) − S12 S21

Y22 =

∗ (Z01

∗ (Z01

∗ (Z01 + S11 Z01 )(1 − S22 ) + S12 S21 Z01 ∗ + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02

S11 =

A=

∗ (Z01 + S11 Z01 )(1 − S22 ) + S12 S21 Z01 √ 2S21 R01 R02

S12

B=

∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ 2S21 R01 R02

C=

(1 − S11 )(1 − S22 ) − S12 S21 √ 2S21 R01 R02

D=

∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ 2S21 R01 R02

S21

∗ ∗ AZ02 + B − CZ01 Z02 − DZ01 AZ02 + B + CZ01 Z02 + DZ01 √ 2(AD − BC)( R01 R02 ) = AZ02 + B + CZ01 Z02 + DZ01 √ 2 R01 R02 = AZ02 + B + CZ01 Z02 + DZ01

S22 =

∗ ∗ −AZ02 + B − CZ01 Z02 + DZ01 AZ02 + B + CZ01 Z02 + DZ01

∗ (h11 − Z01 )(1 + h22 Z02 ) − h12 h21 Z02 (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02 √ 2h12 R01 R02 = (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02 √ −2h21 R01 R02 = (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02

∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) + S12 S21 Z01 Z02 ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ 2S12 R01 R02 = ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ −2S21 R01 R02 = ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02

S11 =

h11 =

S12

h12

S21

S22 =

or

∗ ∗ (Z01 + h11 )(1 − h22 Z02 ) + h12 h21 Z02 (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02

h21

h22 =

(1 − S11 )(1 − S22 ) − S12 S21 ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02

[Z] = {[U ] + [S]}{[U ] − [S]}−1

(8.7.7)

Similarly, other conversion relations can be developed. Table 8.4 lists these equations when the two ports have different complex characteristic impedances. Characteristic impedance at port 1 is assumed Z01 , with its real part as R01 . Similarly, characteristic impedance at port 2 is Z02 , with its real part as R02 .

325

SUGGESTED READING

TABLE 8.5 Conversion Between Scattering and Chain Scattering Parameters S11 =

T21 T11

T11 =

S12 =

T11 T22 − T12 T21 T11

T12

S21 =

1 T11

T21 =

S11 S21

S22 =

−T12 T11

T22 =

−(S11 S22 − S12 S21 ) S21

1 S21 −S22 = S21

8.8 CHAIN SCATTERING PARAMETERS Chain scattering parameters, also known as scattering transfer parameters or T parameters, are useful for networks in cascade. These are defined on the basis of waves a1 and b1 at port 1 as dependent variables and waves a2 and b2 at port 2 as independent variables. Hence,

a1 b1

=

T11 T21

T12 T22

b2 a2

Conversion relations of chain scattering parameters with others can easily be developed. Conversions between S- and T-parameters are listed in Table 8.5.

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Gonzalez, G., Microwave Transistor Amplifiers. Upper Saddle River, NJ: Prentice Hall, 1997. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P. A., Microwave Engineering. Englewood Cliffs, NJ: Prentice Hall, 1988.

326

TWO-PORT NETWORKS

PROBLEMS 8.1. Determine the Z and Y matrixes for the two-port network shown in Figure P8.1. The signal frequency is 500 MHz. I1

I2 4 pF

V1

V2

Figure P8.1

8.2. Determine the Z-parameters of the two-port network shown in Figure P8.2. 20 Ω 10 Ω

10 Ω

1Ω

Figure P8.2

8.3. Determine the Z-parameters of the two-port network shown in Figure P8.3 operating at 800 MHz. 50 Ω

Port 1

50 Ω

10 pF

Figure P8.3

Port 2

327

PROBLEMS

8.4. Determine the Z-matrix for the two-port network shown in Figure P8.4. j2 Ω

j3 Ω j4 Ω

Figure P8.4

8.5. The two-port network shown in Figure P8.5 represents a high-frequency equivalent model of the transistor. If the transistor is operating at 108 rad/s, find its Z-parameters. 1 pF

V1

5 pF 100 kΩ

10 kΩ

0.01V1

V2

Figure P8.5

8.6. Determine the Y-parameters of the two-port network shown in Figure P8.6. 50 Ω

Z0 = 50 Ω

Z0 = 50 Ω

50 Ω

Figure P8.6

8.7. Find the Y-parameters of the two-port network shown in Figure P8.7. 3Ω

Port 1

5Ω

10 Ω

Figure P8.7

2Ω

20 Ω

Port 2

328

TWO-PORT NETWORKS

8.8. Measurements are performed with a two-port network as shown in Figure P8.8. The voltages and currents observed are tabulated in Table P8.8. However, there are a few blank entries in this table. Fill in those blanks and find the Y and Z parameters of this two-port network. I1

I2

~ V1

~

V2

Figure P8.8

TABLE P8.8 Experiment

V1 (V)

V2 (V)

1 2 3 4 5

20 50

0 100

100

50

I1 (A)

I2 (A)

4 −20 5

−8 −5 0

5

15

8.9. The Y-matrix of a two-port network is given as

10 −5 [Y ] = 50 20

mS

It is connected in a circuit as shown in Figure P8.9. Find the voltages V1 and V2 . 25 Ω 100 V ∠0°

V1

[Y ] = 10 −5 50 20

mS

V2

100 Ω

Figure P8.9

8.10. Determine the transmission and hybrid parameters for the networks shown in Figure P8.10.

329

PROBLEMS 20 Ω

200 Ω

200 Ω

(a)

10 Ω

10 Ω 200 + j 8 Ω

(b)

Figure P8.10

8.11. Determine the transmission parameters (ABCD) of the circuit shown in Figure P8.11 operating at ω = 106 rad/s. 1 µH

1 µF I2

I1

V1

1Ω

V2

Figure P8.11

8.12. Find the transmission matrix of the two-port network shown in Figure P8.12.

j50 Ω 50 Ω

50 Ω

Figure P8.12

330

TWO-PORT NETWORKS

8.13. Find the transmission matrix (ABCD parameters) of the circuit shown in Figure P8.13 at 4 GHz. 0.159 pF

100 Ω

220 Ω

220 Ω

Figure P8.13

8.14. Find the transmission matrix of the two-port network shown in Figure P8.14. −j50 Ω

50 Ω

j50 Ω

Port 1

Port 2

Figure P8.14

8.15. Determine the S-parameters for the network shown in Figure P8.15.

+j50 Ω

−j50 Ω Z0 = 50 Ω

50 Ω

Z0 = 50 Ω

Figure P8.15

8.16. Calculate the Z and S parameters for the circuit shown in Figure P8.16 at 4 GHz. Both ports of the network have characteristic impedance of 50 . 0.5 pF

292 Ω

17.6 Ω

Figure P8.16

292 Ω

331

PROBLEMS

8.17. Determine the S-parameters of the two-port T-network shown in Figure P8.17. 10 Ω

Z0 = 50 Ω

20 Ω

Z0 = 50 Ω

30 Ω

Figure P8.17

8.18. Determine the S-parameters of the two-port network shown in Figure P8.18.

1Ω

1Ω Z0 = 50 Ω

Z0 = 50 Ω 10 Ω

Figure P8.18

8.19. Determine the S-parameters of the two-port network shown in Figure P8.19. 50 Ω

Z0 = 50 Ω

50 Ω

Z0 = 50 Ω

Figure P8.19

8.20. Determine the S-parameters of the two-port π-network shown in Figure P8.20. 200 Ω

Z0 = 50 Ω

200 Ω

Figure P8.20

200 Ω

Z0 = 50 Ω

332

TWO-PORT NETWORKS

8.21. Scattering variables measured at a port are found as follows: a = 8 − j 5 and b = 2 + j 5. Normalizing impedance is Z0 = 50 . Calculate the corresponding voltage and current. 8.22. Scattering variables measured at a port are a = 0.85 45◦ and b = 0.25 65◦ . The normalizing impedance is Z0 = 50 . Calculate the corresponding voltage and current. 8.23. Calculate the power flow into a port in each case for the following sets of scattering variables at the port: (a) a = 3 + j 5, b = 0.2 − j 0.1 (b) a = 40 − j 30, b = 5 − j 5 8.24. Calculate the power flow into a port in each case for the following sets of scattering variables at the port: (a) a = 5 60◦ , b = 2 −45◦ (b) a = 50 0◦ , b = 10 0◦ 8.25. Calculate the scattering variables a and b in each of the following cases: (a) V = 4 − j 3 V, I = 0.3 + j 0.4 A (b) V = 10 −30◦ V, I = 0.7 70◦ A Assume that characteristic impedance at the port is 100 . 8.26. Find the reflection coefficient for each of the following sets of scattering variables: (a) a = 0.3 + j 0.4, b = 0.1 − 0.2 (b) a = −0.5 + j 0.2, b = 0 − j 0.1 (c) a = 0.5 −70◦ , b = 0.3 20◦ (d) a = 5 0◦ , b = 0.3 90◦ 8.27. The scattering parameters of a two-port network are given as follows: S11 = 0.687 107◦ , S21 = 1.72 −59◦ , S12 = 0.114 −81◦ , and S22 = 0.381 153◦ . A source of 3 mV with an internal resistance of 50 is connected at port 1. Assuming the characteristic impedance at its ports as 50 , determine the scattering variables a1 , b1 , a2 , and b2 for the following load conditions: (a) 50 and (b) 100 .

9 FILTER DESIGN

A circuit designer frequently requires filters to extract the desired frequency spectrum from a wide variety of electrical signals. If a circuit passes all signals from dc through a frequency ωc but stops the rest of the spectrum, it is known as a low-pass filter. The frequency ωc is called its cutoff frequency. Conversely, a high-pass filter stops all signals up to ωc and passes those at higher frequencies. If a circuit passes only a finite frequency band that does not include zero (dc) and infinite frequency, it is called a bandpass filter. Similarly, a bandstop filter passes all signals except a finite band. Thus, bandpass and bandstop filters are specified by two cutoff frequencies to set the frequency band. If a filter is designed to block a single frequency, it is called a notch filter. The ratio of the power delivered by a source to a load with and without a two-port network inserted in between is known as the insertion loss of that twoport. It is generally expressed in decibels. The fraction of the input power that is lost due to reflection at its input port is called the return loss. The ratio of the power delivered to a matched load to that supplied to it by a matched source is called the attenuation of that two-port network. Filters have been designed using active devices such as transistors and operational amplifiers, as well as with only passive devices (inductors and capacitors only). Therefore, these circuits may be classified as active or passive filters. Unlike passive filters, active filters can amplify the signal besides blocking the undesired frequencies. However, passive filters are economical and easy to design. Further, passive filters perform fairly well at higher frequencies. In this chapter we present the design procedure of these passive circuits. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

333

334

FILTER DESIGN

There are two methods available to synthesize passive filters. One of them is known as the image parameter method and the other as the insertion-loss method. The former provides a design that can pass or stop a certain frequency band, but its frequency response cannot be shaped. The insertion-loss method is more powerful in the sense that it provides a specified response of the filter. Both of these techniques are described in this chapter. The chapter concludes with a design overview of microwave filters.

9.1 IMAGE PARAMETER METHOD Consider the two-port network shown in Figure 9.1. V1 and V2 represent voltages at its ports. Currents I1 and I2 are assumed as indicated in the figure. Note that I1 is entering port 1 while I2 is leaving port 2. Further, Zin is the input impedance at port 1 when Zi2 terminates at port 2. Similarly, Z0 is the output impedance with Zi1 connected at port 1. Zi1 and Zi2 are known as the image impedance of the network. Following the transmission parameter description of the two-port, we can write V1 = AV2 + BI2

(9.1.1)

I1 = CV2 + DI2

(9.1.2)

and

Therefore, the impedance Zin at its input can be found as Zin =

V1 AV2 + BI2 AZi2 + B = = I1 CV2 + DI2 CZi2 + D

(9.1.3)

Alternatively, (9.1.1) and (9.1.2) can be rearranged as follows after noting that AD —BC must be unity for a reciprocal network. Hence, V2 = DV1 − BI1

(9.1.4)

I2 = −CV1 + AI1

(9.1.5)

and

I2

Zi1

V1

Zin

A

B

C

D

Z0

Figure 9.1 Two-port network with terminations.

V2

Zi2

IMAGE PARAMETER METHOD

335

Therefore, the output impedance Z0 is found to be Z0 = −

V2 DV1 − BI1 DZi1 + B =− = I2 −CV1 + AI1 CZi1 + A

Note that Zi1 = −

(9.1.6)

V1 I1

For Zin = Zi1 and Z0 = Zi2 , equations (9.1.3) and (9.1.6) give Zi1 (CZi2 + D) = AZi2 + B

(9.1.7)

Zi2 (CZi1 + A) = DZi1 + B

(9.1.8)

and

Subtracting equation (9.1.8) from (9.1.7), we find that Zi2 =

D Zi1 A

(9.1.9)

Now, substituting (9.1.9) into (9.1.7), we have Zi1 =

AB CD

(9.1.10)

Similarly, substituting (9.1.10) into (9.1.9) yields Zi2 =

BD AC

(9.1.11)

The transfer characteristics of the network can be formulated as follows. From (9.1.1), V1 I2 B =A+B =A+ V2 V2 Zi2 or V1 = V2

√ A √ ( AD + BC) D

For a reciprocal two-port network, AD − BC is unity. Therefore, V2 = V1

D AD − BC = √ √ A AD + BC

√ D √ ( AD − BC) A

(9.1.12)

336

FILTER DESIGN

Similarly, from (9.1.2), I2 = I1

√ A √ ( AD − BC) D

(9.1.13)

Note that equation (9.1.12) is similar to (9.1.13) except that the multiplying coefficient in one is the reciprocal of the other. This coefficient may be interpreted as the transformer turns ratio. It is unity for symmetrical T and π networks. The propagation factor γ (equal to α + j β, as usual) of the network can be defined as e−γ =

√

AD −

√

BC

Since AD − BC = 1, we find that cosh γ =

√

AD

(9.1.14)

The characteristic parameters of π and T networks are summarized in Table 9.1; the corresponding low-pass and high-pass circuits are given in Table 9.2.

TABLE 9.1

Parameters of T and π Networks π-Network

T-Network Z1/2

Z1

A =1+

Z1 2Z2

B = Z1 C =

Z2

2Z2

2Z2

ABCD parameters

Z1/2

Z1 2Z2

B = Z1 +

1 Z1 + Z2 4Z22

D =1+

A =1+

Z1 2Z2

C =

1 Z2

D =1+

Z1 Z2 Z1 Z2 = 1 + Z1 /4Z2 ZiT

Image impedance

Ziπ =

Propagation constant, γ

cosh γ = 1 +

Z1 2Z2

ZiT =

Z12 4Z2

Z1 2Z2

Z1 Z1 Z2 1 + 4Z2

cosh γ = 1 +

Z1 2Z2

337

IMAGE PARAMETER METHOD

TABLE 9.2

Constant-k Filter Sections

Filter Type Low-pass

π-Section

T-Section L /2

L

L /2

C/2

C

High-pass

2C

C/2

C

2C

L

2L

2L

For a low-pass T-section as illustrated in Table 9.2, Z1 = j ωL

(9.1.15)

1 j ωC

(9.1.16)

and Z2 =

Therefore, its image impedance can be found from Table 9.1 as follows: ZiT =

L ω2 LC 1− C 4

(9.1.17)

In the case of a dc signal, the second term inside the parentheses will be zero and the resulting image impedance is generally known as the nominal impedance, Z0 . Hence, L Z0 = C Note that the image impedance goes to zero if ω2 LC/4 = 1. The frequency that satisfies this condition is known as the cutoff frequency, ωc . Hence, ωc = √

2 LC

(9.1.18)

338

FILTER DESIGN

Similarly, for a high-pass T-section, 1 j ωC

(9.1.19)

Z2 = j ωL

(9.1.20)

Z1 = and

Therefore, its image impedance is found to be ZiT =

L 1 1− C 4ω2 LC

(9.1.21)

The cutoff frequency of this circuit will be given as 1 ωc = √ 2 LC

(9.1.22)

Example 9.1 Design a low-pass constant-k T-section that has a nominal impedance of 75 and a cutoff frequency of 2 MHz. Plot its frequency response in the frequency band 100 kHz to 10 MHz. SOLUTION Since the nominal impedance Z0 must be 75 , Z0 = 75 =

L C

The cutoff frequency ωc of a low-pass T-section is given by (9.1.18). Therefore, ωc = 2π × 2 × 106 = √

2 LC

These two equations can be solved for the inductance L and capacitance C, as follows: L = 11.9366 µH and C = 2.122 nF This circuit is illustrated in Figure 9.2. Note that inductance L calculated here is twice the value needed for a T-section. Propagation constant γ of this circuit is

339

IMAGE PARAMETER METHOD

5.9683 µH

5.9683 µH

2.122 nF

Figure 9.2

Low-pass constant-k T-section.

0.1

0.2

0.5

1

2

5

10

0.1

0.2

0.5 1 2 Frequency (MHz)

5

10

0.1

0.2

0.5

2

5

10

0.2

0.5 1 2 Frequency (MHz)

5

10

0

Magnitude (dB)

−10

−20

−30

−40

1

−25

Phase (deg)

−50 −75 −100 −125 −150 −175 0.1

Figure 9.3

Frequency response of the T-section shown in Figure 9.2.

determined from the formula listed in Table 8.1. The transfer characteristics are then found as e−γ . The frequency response of the designed circuit is shown in Figure 9.3. The magnitude of the transfer function (ratio of the output to input voltages) remains constant at 0 dB for frequencies lower than 2 MHz. Therefore, the

340

FILTER DESIGN

output magnitude will be equal to the input in this range. It falls by 3 dB if the signal frequency approaches 2 MHz. It falls continuously as the frequency increases further. Note that the phase angle of the transfer function remains constant at −180◦ beyond 2 MHz. However, it shows a nonlinear characteristic in the passband. The image impedance ZiT of this T-section can be found from (9.1.17). Its characteristics (magnitude and phase angle) with frequency are displayed in Figure 9.4. The magnitude of ZiT reduces continuously as frequency is increased and becomes zero at the cutoff. The phase angle of ZiT is zero in passband, and it changes to 90◦ in the stopband. Thus, image impedance is a variable resistance in the passband, whereas it switches to an inductive reactance in the stopband. The frequency characteristics illustrated in Figures 9.3 and 9.4 are representative of any constant-k filter. There are two major drawbacks to this type of filter: 1. The signal attenuation rate after the cutoff point is not very sharp, 2. The image impedance is not constant with frequency. From a design point of view, it is important that it stay constant, at least in its passband.

350

Magnitude (Ω)

300 250 200 150 100 50 2

4 6 Frequency (MHz)

8

10

80 60 40 20

2

Figure 9.4 frequency.

4 6 Frequency (MHz)

8

10

Image impedance of the constant-k filter of Figure 9.2 as a function of

341

IMAGE PARAMETER METHOD

These problems can be remedied using the techniques described in the following section.

m-Derived Filter Sections Consider the two T-sections shown in Figure 9.5. The first network represents the constant-k filter considered in the preceding section, and the second is a new m-derived section. It is assumed that the two networks have the same image impedance. From Table 9.1 we can write ZiT

= ZiT =

Z1 Z2

Z 1 + 1 4Z2

=

Z1 Z1 Z2 1 + 4Z2

It can be solved for Z2 as follows: Z2

1 = Z1

Now assume that

Z 2 − Z12 Z1 Z2 + 1 4

(9.1.23)

Z1 = mZ1

(9.1.24)

Substituting (9.1.24) in to (9.1.23), we get Z2 =

1 − m2 Z2 + Z1 m 4m

(9.1.25)

Thus, an m-derived section is designed from the values of components determined for the corresponding constant-k filter. The value of m is selected to sharpen the attenuation at cutoff or to control the image impedance characteristics in the passband.

Z1/2

Figure 9.5

Z1/2

Z′1/2

Z′1/2

Z2

Z′2

(a)

(b)

Constant-k T-section (a) and m-derived section (b).

342

FILTER DESIGN

For a low-pass filter, the m-derived section can be designed from the corresponding constant-k filter using (9.1.24) and (9.1.25) as follows: Z1 = j ωmL

(9.1.26)

and Z2 =

1 1 − m2 j ωL + 4m j ωmC

(9.1.27)

Now we need to find its propagation constant γ and devise some way to control its attenuation around the cutoff. Expression for the propagation constant of a Tsection is listed in Table 9.1. To find γ of this T-section, we first divide (9.1.26) by (9.1.27): Z1 ω2 m2 LC 4ω2 m2 /ω2c = − = − Z2 1 − [(1 − m2 )/4]ω2 LC 1 − (1 − m2 )ω2 /ω2c where ωc = √

2 LC

(9.1.28)

(9.1.29)

Using the formula listed in Table 9.1 and (9.1.28), the propagation constant γ is found as follows: cosh γ = 1 +

2 (mω/ωc )2 Z1 =1− 2Z2 1 − (1 − m2 ) (ω/ωc )2

or cosh γ =

ω2c − ω2 − (mω)2 ω2c − (1 − m2 )ω2

(9.1.30)

Hence, the right-hand side of (9.1.30) is dependent on the frequency ω. It will go to infinity (and therefore, γ) if the following condition is satisfied: ω= √

ωc 1 − m2

= ω∞

(9.1.31)

This condition can be used to sharpen the attenuation at cutoff. A small m will place ω∞ close to ω. In other words, ω∞ is selected a little higher than ωc and the fractional value of m is then determined from (9.1.31). Z1 and Z2 of the mderived section are subsequently found using (9.1.24) and (9.1.25), respectively. Note that the image impedance of an m-derived T-section is the same as that of the corresponding constant-k network. However, it will be a function of m in the case of a π-network. Therefore, this characteristic can be utilized to design input and output networks of the filter so that the image impedance of the composite circuit stays almost constant in its passband. Further, an infinite cascade of T-sections can be considered as a π-network after splitting its shunt

343

IMAGE PARAMETER METHOD

Z1′/2

Z1′/2

Z1′/2

Z2′

Z1′/2

Z2′

(a) Z1′

2Z2′

2Z2′

(b)

Figure 9.6 Deembedding of a π-network (b) from a cascaded T-network (a).

arm as illustrated in Figure 9.6. Note that the impedance Z2 of the T-network is replaced by 2Z2 , while two halves of the series arms of the T-network give Z1 of the π-network. Image impedance Ziπ is found from Table 9.1 as follows: Ziπ =

Z1 Z2 Z1 Z2 + [(1 − m2 )/4]Z12 = ZiT ZiT

(9.1.32)

For the low-pass constant-k filter, we find from (9.1.15) to (9.1.18) that Z1 Z2 =

L = Z02 C

Z12 = −ω2 L2 = and,

ZiT = Z0 1 −

2Z0 ω ωc

ω ωc

2

2

Therefore, Ziπ =

1 − (1 − m2 ) (ω/ωc )2 Z0 1 − (ω/ωc )2

344

FILTER DESIGN 2.5

Normalized impedance

2.0

m = 0.8

1.5

m = 0.6

1.0 m = 0.4 0.5 0.0

0.2

0.4

0.6

0.8

1.0

Normalized frequency

Figure 9.7 Normalized image impedance of π-network versus normalized frequency for three values of m.

or Z iπ =

1 − (1 − m2 )ω2 (1 − ω2 )

(9.1.33)

where Z iπ = Ziπ /Z0 may be called the normalized image impedance of a πnetwork and ω = ω/ωc the normalized frequency. Equation (9.1.33) is displayed graphically in Figure 9.7. The normalized image impedance is close to unity for nearly 90% of the passband if m is kept around 0.6. Therefore, it can be used as pre- and poststages of the composite filter (after bisecting it) to control the input impedance. These formulas will be developed later in this section. Example 9.2 Design an m-derived T-section low-pass filter with a cutoff frequency of 2 MHz and a nominal impedance of 75 . Assume that f∞ = 2.05 MHz. Plot the response of this filter in the frequency band of 100 kHz to 10 MHz. SOLUTION From (9.1.31), 1 − m2 = Therefore,

m=

1−

fc f∞

fc f∞

2 =

1−

2

2 2.05

2 = 0.2195

345

IMAGE PARAMETER METHOD

TABLE 9.3

Design Relations for Composite Filters

Low-Pass

High-Pass

Constant-k T-filter

Constant-k T-filter Z0 =

L/2

L/2 C

√

L/C √ ωc = 2/ LC L = 2Z0 /ωc C = 2/Z0 ωc

Z0 = 2C

L

2C

√

L/C √ ωc = 1/2 LC L = 0.5Z0 /ωc C = 0.5/Z0 ωc

m-derived T-Section

m-derived T-Section

(Values of L and C are the same as above) fc 2 m= 1− f∞

(Values of L and C are the same as above) f∞ 2 m= 1− fc

mL/2

mL/2

2C/m

1 − m2 L 4m

L/m 4m C 1 − m2

mC

Input and Output Matching Sections

0.3L 8 L 15 0.3C

2C/m

Input and Output Matching Sections

0.3L 8L 15 0.3C

C/0.3 L/0.3 15 C 8

C/0.3 L/0.3 15 C 8

Using (9.1.24) and (9.1.25) or Table 9.3 and component values of the constantk section obtained earlier in Example 9.1, the m-derived filter is designed as follows: mL/2 = 0.2195 × 5.9683 = 1.31 µH mC = 0.2195 × 2.122 = 465.78 nF and

1 − m2 L = 12.94 µH 4m

This filter circuit is illustrated in Figure 9.8.

346

FILTER DESIGN 1.31 µH

1.31 µH

12.94 µH 465.78 pF

Figure 9.8

An m-derived T-section for Example 9.2.

As noted earlier, the image impedance of this circuit will be the same as that of the corresponding constant-k section. Hence, it will vary with frequency as illustrated in Figure 9.4. The propagation constant γ of this circuit is determined from the formula listed in Table 9.1. The transfer characteristics are then found as e−γ . Its magnitude and phase characteristics versus frequency are illustrated in Figure 9.9. The transfer characteristics illustrated in Figure 9.9 indicate that the m-derived filter has a sharp change at a cutoff frequency of 2 MHz. However, the output signal rises to −4 dB in its stopband. On the other hand, the constant-k filter provides higher attenuation in its stopband. For example, the m-derived filter characteristic in Figure 9.9 shows only 4-dB attenuation at 6 MHz, whereas the corresponding constant-k T-section has an attenuation of more than 30 dB at this frequency, as depicted in Figure 9.3. Composite Filters As demonstrated through the preceding examples, the m-derived filter provides an infinitely sharp attenuation right at its cutoff. However, the attenuation in its stopband is unacceptably low. Contrary to this, the constant-k filter shows a higher attenuation in its stopband, although the change is unacceptably gradual at the transition from passband to stopband. One way to solve the problem is to cascade these two filters. Since image impedance stays the same in two cases, this cascading will not create a new impedance-matching problem. However, we still need to address the problem of image impedance variation with frequency at the input and output ports of the network. As illustrated in Figure 9.7, the image impedance of the π-section with m = 0.6 remains almost constant over 90% of the passband. If this network is bisected to connect on either side of the cascaded constant-k and m-derived sections, it should provide the desired impedance characteristics. To verify these characteristics, let us consider a bisected π-section as shown in Figure 9.10. Its transmission parameters can be found easily following the procedure described in Chapter 8. From (8.4.4) to (8.4.7), we find that A=1+

Z1 4Z2

(9.1.34)

347

IMAGE PARAMETER METHOD 0.1

0.2

0.5

1

2

5

10

0.2

0.5

1

2

5

10

2

5

10

5

10

0

Magnitude (dB)

−2 −4 −6 −8 −10 0.1

Frequency (MHz) 0.1

0.2

0.5

1

0.1

0.2

0.5

0 −25

Phase (deg)

−50 −75 −100 −125 −150 −175 1

2

Frequency (MHz)

Figure 9.9 Frequency response of the m-derived T-section shown in Figure 9.8.

Z′1/2

Zi1

Figure 9.10

2Z′2

Zi2

Right-hand side of the bisected π-section shown in Figure 9.6.

348

FILTER DESIGN

Z1 2 1 C= 2Z2

B=

(9.1.35) (9.1.36)

and D=1

(9.1.37)

Image impedance Zi1 and Zi2 can now be found from (9.1.10) and (9.1.11) as follows: Z2 (9.1.38) Zi1 = Z1 Z2 + 1 = ZiT 4 and

Zi2 =

Z1 Z2 Z1 Z2 = Ziπ = 1 + Z1 /4Z2 ZiT

(9.1.39)

Thus, the bisected π-section can be connected at the input and output ports of cascaded constant-k and m-derived sections to obtain a composite filter that solves the impedance problem as well. Relations for the design of these circuits are summarized in Table 9.3. Example 9.3 Design a low-pass composite filter with a cutoff frequency of 2 MHz and an image impedance of 75 in its passband. Assume that f∞ = 2.05 MHz. Plot its response in the frequency range 100 kHz to 10 MHz. SOLUTION Constant-k and m-derived sections of this filter are designed in Examples 9.1 and 9.2, respectively. The filter’s input and output matching sections can be designed as follows. With m = 0.6, we find that mL = 3.581 µH 2 mC = 636.6 pF 2 and 1 − m2 L = 6.366 µH 2m This composite filter is depicted in Figure 9.11. Figure 9.12 illustrates the frequency response of this composite filter. As it indicates, there is a fairly sharp change in output signal as the frequency changes from its passband to the stopband. At the same time, the output stays well below −40 dB in its stopband. Figure 9.13 shows variation in the image impedance

349

IMAGE PARAMETER METHOD

1.31 µH

3.581 µH 6.366 µH

1.31 µH

5.9683 µH 5.9683 µH

3.581 µH

12.94 µH

6.366 µH

465.78 pF

636.6 pF

2.122 nF 636.6 pF

Bisected Π

Constant-k

Bisected Π

m-derived

Figure 9.11 Composite filter of Example 9.3. 0.1

0.2

0.5

1

2

5

10

0.2

0.5

1

2

5

10

2

5

10

0

Magnitude (dB)

−10 −20 −30 −40 −50 0.1

Frequency (MHz) 0.1

0.2

0.5

1

0.2

0.5

1

150

Phase (deg)

100 50 0 −50 −100 −150 0.1

2

5

10

Frequency (MHz)

Figure 9.12 Frequency response of the composite filter of Figure 9.11.

350

FILTER DESIGN

200

Magnitude (Ω)

150

100

50

1

2 Frequency (MHz)

3

4

75

Phase (deg)

50 25 1 −25

2

3

4

Frequency (MHz)

−50 −75

Figure 9.13 Image impedance of the composite filter of Figure 9.11 versus frequency.

of this filter as the signal frequency changes. This indicates that the image impedance stays at 75 (pure real, because the phase angle is zero) over most of its passband. Example 9.4 Design a high-pass composite filter with a nominal impedance of 75 . It must pass all signals over 2 MHz. Assume that f∞ = 1.95 MHz. Plot its characteristics in the frequency range 1 to 10 MHz. SOLUTION From Table 9.3, we find the components of its constant-k section as follows: L=

75 H = 2.984 µH 2 × 2 × π × 2 × 106

C=

1 F = 530.5 pF 2 × 2 × π × 2 × 106 × 75

and

351

IMAGE PARAMETER METHOD

Similarly, the component values for its m-derived filter section are determined as follows: f∞ 2 1.95 2 m= 1− = 1− = 0.222 fc 2 Hence, 2C = 4.775 nF m L = 13.43 µH m and 4m C = 0.496 nF 1 − m2 The component values for the bisected π-section to be used at its input and output ports are found as C = 1.768 nF 0.3 L = 9.947 µH 0.3 and 15 C = 0.9947 nF 8 The composite filter (after simplifying for the series capacitors in various sections) is illustrated in Figure 9.14. The frequency response of this composite filter is depicted in Figure 9.15. It shows that the attenuation in its stopband stays below −40 dB, and the switching to passband is fairly sharp. As usual with this type of circuit, its phase characteristics may not be acceptable for certain applications because of inherent distortion. 0.6631 nF

9.947 µH

0.9947 nF

0.8681 nF

2.984 µH

1.2903 nF

13.43 µH

0.496 nF

9.947 µH

0.9947 nF

Figure 9.14 Composite high-pass filter with 2-MHz cutoff frequency.

352

FILTER DESIGN 1

1.5

2

1

1.5

2

3

5

7

10

3 5 Frequency (MHz)

7

10

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

1

1.5

2

3

1

1.5

2

5

7

3 5 Frequency (MHz)

7

10

150

Phase (deg)

100 50 0 −50 −100 −150 10

Figure 9.15 Frequency response of the composite high-pass filter of Figure 9.14.

Magnitude (Ω)

250 200 150 100 50 5 10 15 Frequency (MHz)

20

5 10 15 Frequency (MHz)

20

Phase (deg)

80 60 40 20

Figure 9.16 Image impedance of the composite high-pass filter of Figure 9.14 as a function of signal frequency.

INSERTION-LOSS METHOD

353

As depicted in Figure 9.16, the image impedance of this composite filter is almost constant at 75 . Note that its magnitude varies over a wide range in the stopband while its phase angle seems to remain constant at 90◦ . The phase angle of the image impedance is zero in the passband.

9.2 INSERTION-LOSS METHOD The output of an ideal filter would be the same as its input in the passband, whereas it would be zero in the stopband. The phase response of this filter must be linear to avoid signal distortion. In reality, such circuits do not exist and a compromise is needed to design the filters. The image parameter method described in Section 9.1 provides a simple design procedure. However, the transfer characteristics of this circuit cannot be shaped as desired. On the other hand, the insertion-loss method provides ways to shape pass- and stopbands of the filter, although its design theory is much more complex. The power-loss ratio of a two-port network is defined as the ratio of the power that is delivered to the load when it is connected directly at the generator to the power delivered when the network is inserted between the two. In other words, power incident at port 1 1 = power delivered to the load connected at port 2 1 − |(ω)|2 (9.2.1) The power-loss ratio, PLR , expressed in decibels, is generally known as the insertion loss of the network. It can be proved that |(ω)|2 must be an even function of ω for a physically realizable network. Therefore, polynomials of ω2 can represent it as follows. PLR =

|(ω)|2 =

f1 (ω2 ) f1 (ω2 ) + f2 (ω2 )

(9.2.2)

f1 (ω2 ) f2 (ω2 )

(9.2.3)

and PLR = 1 +

where f1 (ω2 ) and f2 (ω2 ) are real polynomials in ω2 . Alternatively, the magnitude of the voltage gain of the two-port network can be found as 1 1 |G(ω)| = √ = (9.2.4) PLR 1 + f1 (ω2 )/f2 (ω2 ) Hence, if PLR is specified, (ω) is fixed. Therefore, the insertion-loss method is similar to the impedance-matching methods discussed in Chapter 7. Traditionally, the filter design begins with a lumped-element low-pass network that is synthesized by using normalized tables. It is subsequently scaled to the desired cutoff frequency and the impedance. Also, the low-pass prototype can

354

FILTER DESIGN

G(ω)

1

0

Passband

Stopband ωc

ω

Figure 9.17 Characteristics of an ideal low-pass filter.

be transformed to obtain a high-pass, a bandpass, or a bandstop filter. These lumped-element filters are used as a starting point to design the transmission line filter. In this section we present the design procedure for two different types of lumped-element low-pass filters. It is followed by the transformation techniques used to design high-pass, bandpass, and bandstop filters. Low-Pass Filters As illustrated in Figure 9.17, an ideal low-pass filter will pass the signals below its cutoff frequency ωc without attenuation while it will stop those with higher frequencies. Further, the transition from its passband to its stopband will be sharp. In reality, this type of filter cannot be designed. Several approximations to these characteristics are available that can be physically synthesized. Two of these are presented below. Maximally Flat Filter As its name suggests, this type of filter provides the flattest possible passband response. However, its transition from passband to stopband is gradual. This is also known as a binomial or Butterworth filter. The magnitude of its voltage gain (and hence, its frequency response) is given as follows: 1 |G(ω)| = 1 + ζω2n

n = 1, 2, 3, . . .

(9.2.5)

where ζ is a constant that controls the power-loss ratio at its band edge, n is the order, and ω is the normalized frequency. For |G(ω)| = 0.7071 (i.e., −3 dB) at the band edge, ζ is unity. Note that derivatives of |G(ω)| at normalized frequencies much below the band edge are close to zero. This guarantees maximum flatness in the passband. Further, (9.2.5) indicates that |G(ω)| will be a fractional number for ω greater than zero. Therefore, it is generally known as the insertion loss, L, of the network when expressed in decibels. Hence, 2n 1 ω = 10 log10 1 + ζ dB L = −20 log10 2n ω c 1 + ζ (ω/ωc ) (9.2.6)

355

INSERTION-LOSS METHOD

Normalized frequency 0.2

0.4

0.6

0.8

−0.5

1 n=2

−1 G(ω)(dB)

1.2

n=4

−1.5

n=7

−2 −2.5

Figure 9.18 Characteristics of maximally flat low-pass filters.

2n ω L = log10 1 + ζ 10 ωc

or

or

ω ζ ωc

2n = 10L/10 − 1

(9.2.7)

Figure 9.18 illustrates the passband characteristics of this type of filter for three different values of n. The power-loss ratio at its band edge is assumed to be −3 dB, and therefore ζ is equal to unity. It shows that the passband becomes flatter with a sharper cutoff when its order n is increased. However, this relationship is not a linear one. This change in characteristics is significant for lower values of n. Equation (9.2.7) can be used for determining the ζ and n values of a filter as follows. If the insertion loss at the band edge (ω = ωc ) is specified as L = Lc , (9.2.7) yields ζ = 10Lc /10 − 1 (9.2.8) Similarly, the required order n (and hence, number of elements required) of a filter can be determined for the L specified at a given stopband frequency. It is found as follows: n=

1 log10 (10L/10 − 1) − log10 ζ × 2 log10 (ω/ωc )

(9.2.9)

Chebyshev Filter A filter with a sharper cutoff can be realized at the cost of flatness in its passband. Chebyshev filters possess ripples in the passband but provide a sharp transition

356

FILTER DESIGN

into the stopband. In this case, Chebyshev polynomials are used to represent the insertion loss. Mathematically, 1 |G(ω)| = 1 + ζTm2 (ω)

m = 1, 2, 3, . . .

(9.2.10)

where ζ is a constant, ω is normalized frequency, and Tm (ω) is a Chebyshev polynomial of the first kind and degree m, defined in (7.5.2). Figure 9.19 shows the frequency response of a typical Chebyshev filter for m = 7. It assumes that ripples up to −3 dB in its passband are acceptable. A comparison of this characteristic to that for a seventh-order Butterworth filter shown in Figure 9.18 indicates that it has a much sharper transition from passband to stopband. However, it is achieved at the cost of ripples in its passband. As before, the insertion loss of a Chebyshev filter is found as follows:

1

L = −20 log10 1 + ζTm2 (ω/ωc ) or

= 10 log10 1 +

2 −1 ω 1 + ζ cos m cos 10 log 10 ωc L= ω 10 log10 1 + ζ cosh2 m cosh−1 ωc

ζTm2

ω ωc

0 ≤ ω ≤ ωc (9.2.11) ωc < ω

where ζ = 100.1×Gr − 1

(9.2.12)

Normalized frequency 0.2

0.4

0.6

0.8

1

1.2

−1

G(ω) (dB)

−2 −3 −4 −5 −6 −7

Figure 9.19

Characteristics of a low-pass Chebyshev filter for m = 7.

357

INSERTION-LOSS METHOD

Gr is the ripple amplitude in decibels. The order m (and hence, number of elements) of a Chebyshev filter can be found from its characteristics as follows: m=

cosh−1

(100.1×L − 1)/(100.1×Gr − 1) cosh−1 (ω/ωc )

(9.2.13)

where L is required insertion loss in decibels at a specified frequency ω. Example 9.5 It is desired to design a maximally flat low-pass filter with at least 15 dB attenuation at ω = 1.3 ωc and −3 dB at its band edge. How many elements will be required for this filter? If a Chebyshev filter is used with a 3-dB ripple in its passband, find the number of circuit elements. SOLUTION From (9.2.8), ζ = 10Lc /10 − 1 = 100.3 − 1 = 1 and from (9.2.9) we have n=

1 log10 (10L/10 − 1) − log10 ζ log10 (101.5 − 1) × = 0.5 × = 6.52 2 log10 (ω/ωc ) log10 (1.3)

Therefore, seven elements will be needed for this maximally flat filter. In the case of a Chebyshev filter, (9.2.13) gives m=

cosh−1

√ cosh−1 101.5 − 1 (100.1×L − 1)/(100.1×Gr − 1) = = 3.17 cosh−1 (ω/ωc ) cosh−1 (1.3)

Hence, it will require only three elements. The characteristics of these two filters are illustrated in Figure 9.20.

Normalized frequency

−1

0.5

1

1.5

2

G(ω) (dB)

−2 −3 −4 −5 −6 −7

Figure 9.20 Characteristics of low-pass Butterworth and Chebyshev filters.

358

FILTER DESIGN

Low-Pass Filter Synthesis The scope of this book does not include the theory of passive filter synthesis. Several excellent references are available in the literature for those who may be interested in this. The design procedure presented here is based on a doubly terminated low-pass ladder network, as shown in Figure 9.21. The g notation used in the figure signifies the roots of an nth-order transfer function that govern its characteristics. These represent the normalized reactance values of filter elements with a cutoff frequency ωc = 1. The source resistance is represented by g0 , and the load is gn+1 . The filter is made up of series inductors and shunt capacitors that are in the form of cascaded T-networks. Another possible configuration is the cascaded π-network that is obtained after replacing g1 by a short circuit, connecting a capacitor across the load gn+1 , and renumbering filter elements 1 through n. Elements of this filter are determined from the n roots of the transfer function. The transfer function is selected according to the pass- and stopband characteristics desired. Normalized values of elements are then found from the roots of that transfer function. These values are then adjusted according to the desired cutoff frequency and the source and load resistance. Design procedures for the maximally flat and the Chebyshev filters can be summarized as follows. Assume that the cutoff frequency is given as follows: ωc = 1

(9.2.14)

Butterworth and Chebyshev filters can then be designed using the following formulas. For a Butterworth filter, g0 = gn+1 = 1

(9.2.15)

and gp = 2 sin

(2p − 1)π 2n

p = 1, 2, · · · , n

(9.2.16)

Element values computed from (9.2.15) and (9.2.16) are given in Table 9.4 for n up to 7.

g1

g0

g3

g2

g5

g4

gn−2

gn

gn−1

Figure 9.21 Low-pass ladder network prototype.

gn+1

359

INSERTION-LOSS METHOD

TABLE 9.4 ωc = 1)

Element Values for Low-Pass Binomial Filter Prototypes (g0 = 1,

n

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

2.0000 1.4142 1.0000 0.7654 0.6180 0.5176 0.4450

1.0000 1.4142 2.0000 1.8478 1.6180 1.4142 1.2470

1 1.0000 1.8478 2.0000 1.9319 1.8019

1.0000 0.7654 1.6180 1.9319 2.0000

1.0000 0.6180 1.4142 1.8019

1 0.5176 1.2470

1 0.445

1.0000

For a Chebyshev filter, g0 = 1 1 gm+1 = ξ coth 4

(9.2.17) m is an odd number m is an even number

g1 =

2a1 χ

gp =

4a(p−1) ap b(p−1) g(p−1)

(9.2.18)

(9.2.19)

and p = 2, 3, · · · , m

(9.2.20)

where

Gr ξ = ln coth 17.37 χ = sinh ap = sin

ξ 2m

(9.2.21) (9.2.22)

(2p − 1)π 2m

(9.2.23)

pπ m

(9.2.24)

and bp = χ2 + sin2

The element values for several low-pass Chebyshev filters that are computed from (9.2.17) to (9.2.24) are given in Tables 9.5 to 9.9.

360

FILTER DESIGN

TABLE 9.5 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 0.1 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

0.3053 0.8431 1.0316 1.1088 1.1468 1.1681 1.1812

1.000 0.6220 1.1474 1.3062 1.3712 1.4040 1.4228

1.3554 1.0316 1.7704 1.9750 2.0562 2.0967

1.0000 0.8181 1.3712 1.5171 1.5734

1.3554 1.1468 1.9029 2.0967

1.0000 0.8618 1.4228

1.3554 1.1812

1.0000

TABLE 9.6 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 0.5 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

0.6987 1.4029 1.5963 1.6703 1.7058 1.7254 1.7373

1.0000 0.7071 1.0967 1.1926 1.2296 1.2479 1.2582

1.9841 1.5963 2.3662 2.5409 2.6064 2.6383

1.0000 0.8419 1.2296 1.3136 1.3443

1.9841 1.7058 2.4759 2.6383

1.0000 0.8696 1.2582

1.9841 1.7373

1.0000

TABLE 9.7 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 1.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.0178 1.8220 2.0237 2.0991 2.1350 2.1547 2.1666

1.0000 0.6850 0.9941 1.0644 1.0911 1.1041 1.1115

2.6599 2.0237 2.8312 3.0010 3.0635 3.0937

1.0000 0.7892 1.0911 1.1518 1.1735

2.6599 2.1350 2.9368 3.0937

1.0000 0.8101 1.1115

2.6599 2.1666

1.0000

Scaling the Prototype to the Desired Cutoff Frequency and Load ž

Frequency scaling. For scaling the frequency from 1 to ωc , divide all normalized g values that represent capacitors or inductors by the desired cutoff frequency expressed in radians per second. Resistors are excluded from this operation.

361

INSERTION-LOSS METHOD

TABLE 9.8 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 2.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.5297 2.4883 2.7108 2.7926 2.8311 2.8522 2.8651

1.0000 0.6075 0.8326 0.8805 0.8984 0.9071 0.9120

4.0957 2.7108 3.6064 3.7829 3.8468 3.8776

1.0000 0.6818 0.8984 0.9392 0.9536

4.0957 2.8311 3.7153 3.8776

1.0000 0.6964 0.9120

4.0957 2.8651

1.0000

TABLE 9.9 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 3.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.9954 3.1014 3.3489 3.4391 3.4815 3.5047 3.5187

1.0000 0.5339 0.7117 0.7483 0.7619 0.7685 0.7722

5.8095 3.3489 4.3473 4.5378 4.6063 4.6392

1.0000 0.5920 0.7619 0.7929 0.8038

5.8095 3.4815 4.4643 4.6392

1.0000 0.6033 0.7722

5.8095 3.5187

1.0000

ž

Impedance scaling. To scale g0 and gn+1 to X from unity, multiply all g values that represent resistors or inductors by X. On other hand, divide those g values representing capacitors by X.

Example 9.6 Design a Butterworth filter with a cutoff frequency of 10 MHz and an insertion loss of 30 dB at 40 MHz. It is to be used between a 50- load and a generator with an internal resistance of 50 . SOLUTION From (9.2.9), we have n= =

1 log10 (1030/10 − 1) − log10 (103/10 − 1) × 2 log10 (40/10) 1 log10 (103 − 1) − log10 (1.9953 − 1) 0.5 × 3 × ≈ ≈ 2.5 2 log10 (4) 0.6

Since the number of elements must be an integer, selecting n = 3 will provide more than the attenuation specified at 40 MHz. From (9.2.15) and (9.2.16), g0 = g4 = 1

362

FILTER DESIGN 795.77 nH 50 Ω R0 = Rs

Figure 9.22

795.77 nH

L1

L3 R4 = RL

636.62 pF

C2

50 Ω

Maximally flat low-pass filter obtained in Example 9.6.

π =1 6 π g2 = 2 sin = 2 2 g1 = 2 sin

and g3 = 2 sin

5π =1 6

If we use the circuit arrangement illustrated in Figure 9.22, element values can be scaled to match the frequency and load resistance. Using the two rules, these values are found as follows: L1 = L3 = 50 ×

1 H = 795.77 nH 2π × 107

and C2 =

1 2 F = 636.62 pF × 50 2π × 107

The frequency response of this filter is shown in Figure 9.23. It indicates a 6dB insertion loss in the passband of the filter. This happens because the source resistance is considered a part of this circuit. In other words, this represents voltage across RL with respect to source voltage, not to input of the circuit. Since source resistance is equal to the load, there is equal division of source voltage that results in −6 dB. Another 3-dB loss at the band edge shows a total of about 9 dB at 10 MHz. This characteristic shows that there is an insertion loss of over 40 dB at 40 MHz. It is clearly more than 30 dB from the band edge, as required by the example. Also, there is a nonlinear phase variation in its passband. Alternatively, we can use a circuit topology as shown in Figure 9.24. In that case, component values are found by using the scaling rules: L2 = 50 ×

1 × 2 = 0.15915 × 10−5 H = 1591.5 nH 2π × 107

C1 = C3 =

1 1 10−9 × F = 318.31 pF ×1= 7 50 2π × 10 π

and

363

INSERTION-LOSS METHOD 0.1

0.5

1

0.1

0.5

1

0.1

0.5

1

0.1

0.5

1

5

10

50

100

5 10 Frequency (MHz)

50

100

10

50

100

5 10 Frequency (MHz)

50

100

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

5

150

Phase (deg)

100 50 0 −50 −100 −150

Figure 9.23 Characteristics of the low-pass filter shown in Figure 9.22.

50 Ω R1 = Rs

C1

1.5915 µH L2

318.31 pF

Figure 9.24

R4 = RL

C3 318.31 pF

50 Ω

Alternative circuit for Example 9.6.

The frequency response of this circuit is illustrated in Figure 9.25. It is identical to that shown in Figure 9.23 for the earlier circuit. Example 9.7 Design a low-pass Chebyshev filter that may have ripples no more than 0.01 dB in its passband. The filter must pass all frequencies up to 100 MHz

364

FILTER DESIGN 0.1

0.5

1

0.1

0.5

1

5

10

50

100

5

10

50

100

5

10

50

100

5

10

50

100

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

Frequency (MHz) 0.1

0.5

1

0.1

0.5

1

150

Phase (deg)

100 50 0 −50 −100 −150 Frequency (MHz)

Figure 9.25 Characteristics of the low-pass filter shown in Figure 9.24.

and attenuate the signal at 400 MHz by at least 5 dB. The load and the source resistance are of 75 each. SOLUTION Since Gr = 0.01 dB and

400 L 100

= 5 dB

INSERTION-LOSS METHOD

365

(9.2.13) gives m=

cosh−1

(100.5 − 1)/(100.001 − 1) =2 cosh−1 (4)

Since we want a symmetrical filter with 75 on each side, we select m = 3 (an odd number). It will provide more than 5 dB of insertion loss at 400 MHz. The g values can now be determined from (9.2.17) to (9.2.24), as follows. From (9.2.23), π = 0.5 6 π a2 = sin = 1 2 a1 = sin

and a3 = sin

5π = 0.5 6

Next, from (9.2.21), (9.2.22), and (9.2.24), we get 0.01 ξ = ln coth = 7.5 17.37 7.5 = 1.6019 6 π b1 = 1.60192 + sin2 = 3.316 3 2π b2 = 1.60192 + sin2 = 3.316 3 χ = sinh

and b3 = 1.60192 + sin2

3π = 2.566 3

However, b3 is not needed in further calculations. Finally, from (9.2.17) and (9.2.20), g0 = g4 = 1 2 × 0.5 = 0.62425 1.6019 4 × 0.5 × 1 g2 = = 0.9662 3.316 × 0.62425 g1 =

and g3 =

4 × 1 × 0.5 = 0.62425 3.316 × 0.9662

366

FILTER DESIGN

74.5155 nH 75 Ω R0 = R s

Figure 9.26

74.5155 nH L3

L1 20.5 pF

R4 = R L C2

75 Ω

Low-pass Chebyshev filter circuit obtained in Example 9.7.

For the circuit topology of Figure 9.26, element values are found after applying the scaling rules as follows: L1 = L3 =

75 × 0.62425 H = 74.5155 nH 2π × 108

and C2 =

1 1 × × 0.9662 F = 20.5 pF 75 2π × 108

The frequency response of this filter is illustrated in Figure 9.27. As expected, there is over a 20-dB insertion loss at 400 MHz compared with its passband. Note that the magnitude of ripple allowed is so small that it does not show up in this figure. However, it is present there, as shown on an expanded scale in Figure 9.28. Figure 9.28 shows the passband characteristics of this Chebyshev filter. Since the scale-of-magnitude plot is now expanded, the passband ripple is clearly visible. As expected, the ripple stays between −6.02 and −6.03 dB. Phase variation in this passband ranges from 0◦ to −70◦ . The band edge of this filter can be sharpened further either by using a higher-order filter or by allowing larger magnitudes of the ripple in its passband. The higher-order filter will require more elements because the two are directly related. The next example illustrates that a sharper transition between the bands can be obtained even with a three-element filter if ripple magnitudes up to 3 dB are acceptable. Example 9.8 Reconsider Example 9.7 to design a low-pass filter that exhibits the Chebyshev response with 3-dB ripple in its passband, m = 3, and a cutoff frequency of 100 MHz. The filter must have 75 at both its input and output ports. SOLUTION From (8.2.23), π = 0.5 6 π a2 = sin = 1 2

a1 = sin

367

INSERTION-LOSS METHOD 1

5

10

1

5

10

50

100

500 1000

50

100

500 1000

0

Magnitude (dB)

−10 −20 −30 −40 −50 Frequency (MHz) 1

5

10

50

100

500 1000

1

5

10

50

100

500 1000

150 100

Phase (deg)

50 0 −50 −100 −150

Frequency (MHz)

Figure 9.27 Characteristics of the low-pass Chebyshev filter shown in Figure 9.26.

and a3 = sin

5π = 0.5 6

Now, from (9.2.21), (9.2.22), and (9.2.24),

3 ξ = ln coth 17.37 χ = sinh

= 1.7661

1.7661 = 0.2986 6

368

FILTER DESIGN

1

2

5

1

2

5

10

20

50

100

10

20

50

100

Magnitude (dB)

−6.022 −6.024 −6.026 −6.028 −6.03 Frequency (MHz) 1

2

5

1

2

5

10

20

50

100

10

20

50

100

0 −10

Phase (deg)

−20 −30 −40 −50 −60

Frequency (MHz)

Figure 9.28 Passband characteristics of the filter shown in Figure 9.26.

π = 0.8392 3 2π b2 = 0.29862 + sin2 = 0.8392 3

b1 = 0.29862 + sin2

and b3 = 0.29862 + sin2 However, b3 is not needed.

3π = 0.0892 3

369

INSERTION-LOSS METHOD 0.4 µH 75 Ω

L1

R0 = Rs

Figure 9.29

0.4 µH L3 R4 = RL

15.1 pF

C2

75 Ω

Low-pass Chebyshev filter circuit obtained in Example 9.8.

The g values for the filter can now be determined from (9.2.17) to (9.2.20), as follows: g0 = g4 = 1 2 × 0.5 = 3.349 0.2986 4 × 0.5 × 1 g2 = = 0.7116 3.349 × 0.8392

g1 =

and g3 =

4 × 1 × 0.5 = 3.349 0.7116 × 0.8392

These g values are also listed in Table 9.9 for m = 3. Elements of the circuit shown in Figure 9.29 are determined using the scaling rules, as follows (see Figure 9.30 for the characteristics): L1 = L3 = and C2 =

75 × 3.349 H = 0.4 µH 2π × 108

1 1 × 0.7116 F = 15.1 pF × 75 2π × 108

High-Pass Filter As mentioned earlier, a high-pass filter can be designed by transforming the low-pass prototype. This frequency transformation is illustrated in Figure 9.31. As illustrated, an ideal low-pass filter passes all signals up to the normalized frequency of unity with zero insertion loss, whereas it attenuates higher frequencies completely. On the other hand, a high-pass filter must pass all signals with frequencies higher than its cutoff frequency ωc and stop the signals that have lower frequencies. Therefore, the following frequency transformation formula will transform a low-pass filter to a high-pass filter: ω=−

ωc ω

(9.2.25)

370

FILTER DESIGN 1

10

100

1

10 Frequency (MHz)

100

1

10

100

1

10 Frequency (MHz)

100

−6 −7 Magnitude (dB)

−8 −9 −10 −11 −12 −13

100

Phase (deg)

50 0 −50 −100 −150

Characteristics of the low-pass filter of Example 9.8.

Magnitude

Magnitude

Figure 9.30

1

0

1

ω

1

0

ωc

Figure 9.31 Transformation from a low-pass to a high-pass characteristic.

ω

INSERTION-LOSS METHOD

371

Thus, inductors and capacitors will change their places. Inductors will replace the shunt capacitors of the low-pass filter and capacitors will be connected in series, in place of inductors. These elements are determined as follows: CHP =

1 ωc gL

(9.2.26)

LHP =

1 ωc gC

(9.2.27)

and

Capacitor CHP and inductor LHP can now be scaled as required by the load and source resistance. Example 9.9 Design a high-pass Chebyshev filter with passband ripple magnitude less than 0.01 dB. It must pass all frequencies over 100 MHz and exhibit at least 5 dB of attenuation at 25 MHz. Assume that the load and source resistances are at 75 each. SOLUTION The low-pass filter designed in Example 9.7 provides the initial data for this high-pass filter. With m = 3, gL = 0.62425, and gC = 0.9662, we find from (9.2.26) and (9.2.27) that CHP =

1 F = 2.5495 nF 2π × 108 × 0.62425

and LHP =

2π ×

1 H = 1.6472 nH × 0.9662

108

Now, applying the resistance scaling, we get C1 = C3 =

2.5495 nF = 33.9933 pF = 34 pF 75

and L2 = 75 × 1.6472 = 123.5 nH The resulting high-pass Chebyshev filter is shown in Figure 9.32. Its frequency response is illustrated in Figure 9.33. As before, the source resistance is considered a part of the filter circuit and therefore its passband shows a 6-dB insertion loss. Bandpass Filter A bandpass filter can be designed by transforming the low-pass prototype as illustrated in Figure 9.34. Here, an ideal low-pass filter passes all signals up to

372

FILTER DESIGN

34 pF

75 Ω R0 = Rs

34 pF

C1 L2

Figure 9.32

R4 = RL

C3 123.5 nH

75 Ω

High-pass Chebyshev filter obtained for Example 9.8.

20

50

100

200

500

1000

2000

20

50

100

200

500

1000

2000

−7.5

Magnitude (dB)

−10 −12.5 −15 −17.5 −20 −22.5

Frequency (MHz) 20

50

100

200

500

1000

2000

20

50

100

200

500

1000

2000

150

Phase (deg)

100 50 0 −50 −100 −150

Frequency (MHz)

Figure 9.33

Characteristics of the high-pass Chebyshev filter shown in Figure 9.32.

373

Magnitude

Magnitude

INSERTION-LOSS METHOD

1

−1

0

1

ω

1

0

ωl

ω

ωu

Figure 9.34 Transformation from a low-pass to a bandpass characteristic.

the normalized frequency of unity with zero insertion loss, whereas it attenuates higher frequencies completely. On the other hand, a bandpass filter must pass all signals with frequencies between ωl and ωu and stop the signals that are outside this frequency band. Hence, the following frequency relation transforms the response of a low-pass filter to a bandpass: ω=

ω2 − ω20 1 ωu − ωl ω

where ω0 =

√

ωl × ωu

(9.2.28)

(9.2.29)

This transformation replaces the series inductor of low-pass prototype with an inductor LBP1 and a capacitor CBP1 that are connected in series. The component values are determined as follows: CBP1 =

ωu − ωl ω20 gL

(9.2.30)

LBP1 =

gL ωu − ωl

(9.2.31)

and

Also, the capacitor CBP2 that is connected in parallel with an inductor LBP2 will replace the shunt capacitor of the low-pass prototype. These elements are found as follows: LBP2 =

ωu − ωl ω20 gC

(9.2.32)

CBP2 =

gC ωu − ωl

(9.2.33)

and

These elements need to be scaled further as desired by the load and source resistance.

374

FILTER DESIGN

Example 9.10 Design a bandpass Chebyshev filter that exhibits no more than 0.01-dB ripples in its passband. It must pass signals in the frequency band 10 to 40 MHz with zero insertion loss. Assume that the load and source resistances are at 75 each. SOLUTION From (9.2.29), we have f0 =

fl fu = 107 × 40 × 106 = 20 × 106 Hz

Now, using (9.2.30) to (9.2.33) with gL = 0.62425 and gC = 0.9662 from Example 9.7, we get 2π × 106 (40 − 10) F = 19.122 nF (2π × 20 × 106 )2 × 0.62424 0.62424 = H = 3.3116 nH 2π × 30 × 106

CBP1 = LBP1

LBP2 =

2π × 106 (40 − 10) H = 12.354 nH (2π × 20 × 106 )2 × 0.9662

CBP2 =

0.9662 F = 5.1258 nF 2π × 30 × 106

and

Next, values of elements are determined after applying the load and source resistance scaling. Hence, 19.122 nF = 254.96 pF ≈ 255 pF 75 L1 = L3 = 75 × 3.3116 nH = 0.2484 µH C1 = C3 =

L2 = 75 × 12.354 nH = 0.9266 µH and C2 =

5.1258 nF = 68.344 pF 75

The resulting filter circuit is shown in Figure 9.35, and its frequency response is depicted in Figure 9.36. It illustrates that we indeed have transformed the low-pass prototype to a bandpass filter for the 10- to 40-MHz frequency band. As before, the source resistance is considered a part of the filter circuit, and therefore its passband shows a 6-dB insertion loss. Figure 9.37 shows its passband characteristics with expanded scales. It indicates that the ripple magnitude is limited to 0.01 dB, as desired.

375

INSERTION-LOSS METHOD

75 Ω

255 pF

C1

0.2484 µH

0.2484 µH

L1

L3

0.9266 µH L2

Figure 9.35

0

255 pF C3 75 Ω

68.344 pF C2

Bandpass filter circuit for Example 9.10.

0.1

1

10

100

1000

0.1

1

10 Frequency (MHz)

100

1000

0.1

1

10

100

1000

0.1

1

10 Frequency (MHz)

100

1000

Magnitude (dB)

−20 −40 −60 −80 −100 −120

150

Phase (deg)

100 50 0 −50 −100 −150

Figure 9.36

Characteristics of the bandpass filter shown in Figure 9.35.

Bandstop Filter A bandstop filter can be realized by transforming the low-pass prototype as illustrated in Figure 9.38. Here, an ideal low-pass filter passes all signals up to the normalized frequency of unity with zero insertion loss, whereas it completely

376

FILTER DESIGN 10

15

20

30

10

15

20 Frequency (MHz)

30

10

15

20

30

10

15

20 Frequency (MHz)

30

Magnitude (dB)

−6.022 −6.024 −6.026 −6.028 −6.03

60

Phase (deg)

40 20 0 −20 −40 −60

Figure 9.37 Passband characteristics of the bandpass filter shown in Figure 9.35.

attenuates higher frequencies. On the other hand, a bandstop filter must stop all signals with frequencies between ωl and ωu and pass the signals that are outside this frequency band. Hence, its characteristics are opposite to that of a bandpass filter considered earlier. The following frequency relation transforms the response of a low-pass filter to the bandstop: ω = (ωu − ωl )

ω ω2 − ω20

(9.2.34)

This transformation replaces the series inductor of low-pass prototype with an inductor LBS1 and a capacitor CBS1 that are connected in parallel. Component values are determined as follows: LBS1 =

(ωu − ωl )gL ω20

(9.2.35)

377

−1

1

0

ω

1

Magnitude

Magnitude

INSERTION-LOSS METHOD

1

0

ωl

ω

ωu

Figure 9.38 Transformation from a low-pass to a bandstop characteristic.

and CBS1 =

1 (ωu − ωl )gL

(9.2.36)

Also, capacitor CBS2 , which is connected in series with an inductor LBS2 , will replace the shunt capacitor of low-pass prototype. These elements are found as follows: LBS2 =

1 (ωu − ωl )gC

(9.2.37)

CBS2 =

(ωu − ωl )gC ω20

(9.2.38)

and

These elements need to be further scaled as desired by the load and source resistance. Table 9.10 summarizes these transformations for the low-pass prototype filter. Example 9.11 Design a maximally flat bandstop filter with n = 3. It must stop signals in the frequency range 10 to 40 MHz and pass the rest of the frequencies. Assume that the load and source resistances are at 75 each. SOLUTION From (9.2.29), we have f0 = fl fu = 107 × 40 × 106 = 20 × 106 Hz Now, using (9.2.35) to (9.2.38) with gL = 1 and gC = 2 from Example 9.6, we get 2π × 106 (40 − 10) × 1 H = 11.94 nH (2π × 20 × 106 )2 1 = F = 5.305 nF 6 2π × 10 (40 − 10) × 1

LBS1 = CBS1

378

FILTER DESIGN

TABLE 9.10 Filter Transformations Filter

Circuit Elements

Low-pass gC

gL

High-pass 1 ω c gL

1 ωc gC

Bandpass ωu − ω l ω20 gL

ωu − ω l ω20 gC gL ω u − ωl

gC ωu − ωl

Bandstop

ωu − ωl gL

1 ωu − ω l gC

ω20 ωu − ω l gC ω20 1 ωu − ωl gL

LBS2 =

1 H = 2.653 nH 2π × 106 (40 − 10) × 2

CBS2 =

2π × 106 (40 − 10) × 2 F = 23.87 nF (2π × 20 × 106 )2

and

Values of required elements are determined next following the load and source resistance scaling. Hence, C1 = C3 =

5.305 nF = 70.73 pF 75

379

INSERTION-LOSS METHOD 0.8955 µH 75 Ω

0.8955 µH

L1

L3

C1 70.73 pF 318.3 pF

75 Ω L2

0.2 µH

Figure 9.39

−6

C3 70.73 pF

C2

Bandstop circuit for Example 9.11.

0.1

1

10

100

1000

0.1

1

10 Frequency (rad/s)

100

1000

0.1

1

10

100

1000

0.1

1

10 Frequency (rad/s)

100

1000

Magnitude (dB)

−6.5 −7 −7.5 −8

100

Phase (deg)

50 0 −50 −100 −150

Figure 9.40

Characteristics of the bandstop filter shown in Figure 9.39.

380

FILTER DESIGN

L1 = L3 = 75 × 11.94 nH = 0.8955 µH L2 = 75 × 2.653 nH = 0.1989 µH ≈ 0.2 µH and C2 =

23.87 nF = 318.3 pF 75

The resulting filter circuit is shown in Figure 9.39. Its frequency response depicted in Figure 9.40 indicates that we have transformed the low-pass prototype to a bandstop filter for the 10- to 40-MHz frequency band. As before, the source resistance is considered a part of the filter circuit, and therefore its passband shows a 6-dB insertion loss.

9.3 MICROWAVE FILTERS The filter circuits presented so far in this chapter use lumped elements. However, these may have practical limitations at microwave frequencies. When the signal wavelength is short, distances between the filter components need to be taken into account. Further, discrete components at such frequencies may cease to operate due to associated parasitic elements and need to be approximated with distributed components. As found in Chapter 3, transmission line stubs can be used in place of lumped elements. However, there may be certain practical problems in implementing the series stubs. This section begins with a technique to design a low-pass filter with only parallel connected lines of different characteristic impedance values. It is known as the stepped impedance (or high-Z low-Z) filter. Since this technique works mainly for low-pass filters, the procedure to transform series reactance to shunt that utilizes Kuroda’s identities is summarized next. Redundant sections of the transmission line are used to separate filter elements, and therefore this procedure is known as redundant filter synthesis. Nonredundant circuit synthesis makes use of these sections to improve the filter response as well. Other methods of designing microwave filters include coupled transmission lines and resonant cavities. Impedance transformers, discussed in Chapter 7, are essentially bandpass filters with different impedance at the twoports. Interested readers can find detailed design procedures of all these filters in the references at the end of this chapter.

Low-Pass Filter Design Using Stepped Impedance Distributed Elements Consider a lossless transmission line of Z0 , as shown in Figure 9.41(a). If it is Figure 9.41(b), these two must have the ance matrix of the transmission line was

length d and characteristic impedance equivalent to the T-network shown in same network parameters. The impeddetermined in Example 8.4 as follows:

381

MICROWAVE FILTERS

d Z0, β (a)

Z11 − Z12

Z11 − Z12 Z12

(b)

jXL 2

jXL 2

jB

(c)

Figure 9.41 Lossless transmission line (a), equivalent symmetrical T-network (b), and (c) elements of T-networks.

[Z]line =

−j Z0 cot βd −j

Z0 sin βd

Z0 sin βd −j Z0 cot βd −j

The impedance matrix of the T-network, [Z]T , may be found as Z11 Z12 [Z]T = Z12 Z11

(9.3.1)

(9.3.2)

Therefore, if this T-network represents the transmission line, we can write Z11 = −j Z0 cot βd

(9.3.3)

and Z12 = −j

Z0 sin βd

(9.3.4)

Hence, Z12 is a capacitive reactance. Circuit elements in the series arm of the T-network are found to be inductive reactance. This is found as follows: Z11 − Z12 = −j Z0

βd cos βd − 1 sin (βd/2) = j Z0 = j Z0 tan sin βd cos (βd/2) 2

(9.3.5)

382

FILTER DESIGN

Therefore, the series element of the T-network is an inductive reactance while its shunt element is capacitive, provided that βd < π/2. For βd < π/4, (9.3.4) and (9.3.5) give B≈

βd Z0

(9.3.6)

and XL ≈ Z0 βd

(9.3.7)

Some special cases are as follows: 1. For Z0 very large, (9.3.6) becomes negligible compared with (9.3.7). Therefore, the equivalent T-network (and hence the transmission line) represents an inductor that is given by L=

Z0 βd ω

(9.3.8)

2. For Z0 very small, (9.3.6) dominates over (9.3.7). Therefore, the transmission line is effectively representing a shunt capacitance in this case. It is given by βd (9.3.9) C= ωZ0 Hence, high-impedance sections (Z0 = Zh ) of the transmission line can replace the inductors, and low-impedance sections (Z0 = Zm ) can replace the capacitors of a low-pass filter. In a design, these impedance values must be selected as far apart as possible and the section lengths are determined to satisfy (9.3.8) and (9.3.9). When combined with the scaling rules, electrical lengths of the inductive and capacitive sections are found as (βd)L =

R0 L Zh

rad

(9.3.10)

(βd)C =

Zm C R0

rad

(8.3.11)

and

where L and C are normalized element values (g values) of the filter and R0 is the filter impedance. Example 9.12 Design a three-element maximally flat low-pass filter with its cutoff frequency as 1 GHz. It is to be used between a 50- load and a generator with its internal impedance at 50 . Assume that Zh = 150 and Zm = 30 .

383

MICROWAVE FILTERS L1 = 1

L3 = 1

50 Ω R0 = 1

R4 = 1

C2 = 2

50 Ω

Figure 9.42 Maximally flat low-pass filter with normalized elements’ values for Example 9.12.

68.75° 19.1°

Figure 9.43

19.1°

Maximally flat low-pass filter for Example 9.12.

SOLUTION From (9.2.15) and (9.2.16), g0 = g4 = 1 π g1 = 2 sin = 1 6 π g2 = 2 sin = 2 2 and g3 = 2 sin

5π =1 6

If we use a circuit arrangement as illustrated in Figure 9.42, element values can be scaled for the frequency and the impedance. Using the scaling rules, these values are found as follows (see Figure 9.43): θ1 = θ3 = (βd)L =

1 × 50 ◦ = 0.3333 rad = 19.1 150

and θ2 = (βd)C =

30 × 2 ◦ = 1.2 rad = 68.75 50

Filter Synthesis Using the Redundant Elements As mentioned earlier, transmission line sections can replace the lumped elements of a filter circuit. Richard’s transformation provides the tools needed in such a

384

FILTER DESIGN

design. Further, Kuroda’s identities are used to transform series elements to a shunt configuration that facilitates the design. Richard’s Transformation Richard proposed that open- and short-circuited lines could be synthesized like lumped elements through the following transformation: = tan βd = tan

ωd vp

(9.3.12)

where vp is the phase velocity of signal propagating on the line. Since the tangent function is periodic with a period of 2π, (9.3.12) is a periodic transformation. Substituting in place of ω, the reactance of the inductor L and of the capacitor C may be written as follows: j XL = j L = j L tan βd

(9.3.13)

and j XC = −j

1 1 1 = −j = −j cot βd C C tan βd C

(9.3.14)

Comparison of (9.3.13) and (9.3.14) with the special cases considered in Section 3.2 indicates that the former represents a short-circuited line with its characteristic impedance as L, while the latter is an open-circuited with Z0 as 1/C. Filter impedance is assumed to be unity. To obtain the cutoff of a low-pass filter prototype at unity frequency, the following must hold: = tan βd = 1 → βd =

π λ →d= 4 8

(9.3.15)

Note that the transformation holds only at the cutoff frequency ωc (frequency corresponding to λ), and therefore the filter response will differ from that of its prototype. Further, the response repeats every 4 ωc . These transformations are illustrated in Figure 9.44. These stubs are known as the commensurate lines because of their equal lengths. Kuroda’s Identities As mentioned earlier, Kuroda’s identities facilitate the design of distributed element filters, providing the means to transform the series stubs into shunt, or vice versa, to separate the stubs physically, and to render the characteristic impedance realizable. Figure 9.45 illustrates four identities that are useful in such a transformation of networks. These networks employ the unit element (U.E.), which is basically a λ/8-long line at the cutoff frequency ωc , with specified characteristic

385

MICROWAVE FILTERS

ω-plane

Ω-plane L

Z0 = L

Short

1 C

Open

Z0 =

C

d

Figure 9.44 Distributed inductor and capacitor obtained from Richard’s transformation.

Z1 U.E. n2Z1

U.E. Z2

1 n2Z2

(a) Z1 n2 1 Z2

U.E. Z2

U.E. Z1

n2 (b) 1 n2Z2

1 Z2 U.E. n2Z1

U.E. Z1

n2 :1 (c) U.E. Z1

U.E. Z1

Z1

Z2

n2

n2 (d)

n2 = 1 +

1:n2 Z2 Z1

Figure 9.45 Kuroda’s identities.

impedance. Inductors represent short-circuited stubs and capacitors open-circuited stubs. Obviously, its equivalent in lumped circuit theory does not exist. The first identity is proved below. A similar procedure can be used to verify others. The circuits associated with the first identity can be redrawn as shown in Figure 9.46. A short-circuited stub of characteristic impedance Z1 replaces the

386

FILTER DESIGN Short

d

d Z1

Open

d n2Z2

n2Z1

Z2

d

Figure 9.46 Illustration of Kuroda’s first identity, shown in Figure 9.45.

series inductor that is followed by a unit element transmission line of characteristic impedance Z2 . Its transformation gives a unit element transmission line of characteristic impedance n2 Z1 and an open-circuited stub of characteristic impedance n2 Z2 in place of a shunt capacitor. Following Example 8.10, the transmission matrix of a short-circuited series stub with input impedance ZS can be found as follows. The impedance ZS of the series stub is given as ZS = j Z1 tan βd = j Z1 1 j Z1 A B = C D S 0 1

(9.3.16) (9.3.17)

Following Example 8.13, the transmission matrix of the unit element is found to be 1 j Z2 tan βd cos βd j Z2 sin βd A B = cos βd = j sin βd j tan βd C D U.E. cos βd 1 Z2 Z2 or 1 j Z2 1 A B j (9.3.18) =√ C D U.E. 1 + 2 1 Z2 Since these two elements are connected in cascade, the transmission matrix for the first circuit in Figure 9.46 is found from (9.3.17) and (9.3.18) as follows:

A C

B D

=

a

A C

B D

· S

A C

B D

= √ U.E.

1 1+

2

2 Z1 Z2 j Z2

1−

j (Z1 + Z2 )

1 (9.3.19)

387

MICROWAVE FILTERS

Now for the second circuit shown in Figure 9.46, the transmission matrix for the unit element is

A C

B D

U.E.

1

= cos βd j tan βd

j n2 Z1 tan βd 1

n2 Z 1

= √

1 1 + 2

1

j n2 Z 1

j n2 Z1 1

(9.3.20)

Since the impedance ZP of the shunt stub is given as Zp = −j n2 Z2 cot βd = −j

n2 Z2

(9.3.21)

its transmission matrix may be found, by following Example 8.11, as 1 0 A B = j C D p 1 n2 Z2

(9.3.22)

Hence, the transmission matrix of the circuit shown on right-hand side is found as 2 j n Z 1 1 = √ 1 1 + 2 j 1 1 + 2 n Z1 Z2

A C

B D

=

b

A C

B D

· U.E.

A C

B D

1−

p

2 Z1 Z2

(9.3.23)

On substituting for n2 , it is easy to show that this matrix is equal to that found in (9.3.19). Example 9.13 Design a three-element maximally flat low-pass filter with its cutoff frequency as 1 GHz. It is to be used between a 50- load and a generator with its internal impedance at 50 . SOLUTION A step-by-step procedure to design this filter is as follows: ž Step 1. As before, g values for the filter are found from (9.2.15) and (9.2.16) as

g0 = g4 = 1 π g1 = 2 sin = 1 6 π g2 = 2 sin = 2 2 and g3 = 2 sin

5π =1 6

388

FILTER DESIGN L1 = 1

L3 = 1

50 Ω R0 = 1

R4 = 1

C2 = 2

50 Ω (a)

L1 Z0 = 1

L1 Z0 = 1

L3 Z0 = 1

d

L3 Z0 = 1

d

U.E.

U.E.

Z0 = 1 d

d

C2 Z0 = 0.5 (b)

Z0 = 2

Z0 = 2

Z0 = 0.5 (d)

Z0 = 50 Ω

d U.E.

U.E. Z0 = 2

d

C2 Z0 = 0.5

d

(c)

d d

Z0 = 1

Z0 = 100 Ω Z0 = 100 Ω Z0 = 50 Ω

Z0 = 100 Ω

Z0 = 100 Ω

Z0 = 2 d = λ/8 at ω = 1

Z0 = 25 Ω (e)

Figure 9.47 Maximally flat low-pass filter for Example 9.13.

The circuit is drawn with normalized values as shown in Figure 9.47(a). ž Step 2. Using Richard’s transformation, the filter elements are replaced by stubs. Hence, short-circuited series stubs replace inductors L1 and L3 , while an open-circuited shunt stub replaces the capacitor C2 . This is shown in Figure 9.47(b). ž Step 3. Since the circuit found in step 2 cannot be fabricated in its present form, unit elements are added on its two sides. This is shown in Figure 9.47(c). ž Step 4. Using Kuroda’s identity shown in Figure 9.45(a), series stub and unit element combinations are transformed to an open-circuited shunt stub and the unit element. Since Z1 = Z2 = 1, equivalent capacitance C and the characteristic impedance of the corresponding stub are found to be C = 12 = 0.5 and Z0 = 2. Also, the characteristic impedance of each unit element is found to be 2. This circuit is illustrated in Figure 9.47(d). ž Step 5. For impedance and frequency scaling, all normalized characteristic impedances are now multiplied by 50 , and the stub lengths are selected as λ/8 at 1 GHz. This is shown in Figure 9.47(e).

PROBLEMS

389

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Davis, W. A., Microwave Semiconductor Circuit Design. New York: Van Nostrand Reinhold, 1984. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures. Dedham, MA: Artech House, 1980. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998.

PROBLEMS 9.1. Design a low-pass composite filter with nominal impedance of 50 . It must pass all signals up to 10 MHz. Assume that f∞ = 10.05 MHz. Plot its characteristics in the frequency range 1 to 100 MHz. 9.2. Design a high-pass T-section constant-k filter that has a nominal impedance of 50 and a cutoff frequency of 10 MHz. Plot its frequency response in the frequency band 100 kHz to 100 MHz. 9.3. Design an m-derived T-section high-pass filter with a cutoff frequency of 10 MHz and nominal impedance of 50 . Assume that f∞ = 9.95 MHz. Plot the response of this filter in the frequency band 100 kHz to 100 MHz. 9.4. Design a high-pass composite filter with a cutoff frequency of 10 MHz and image impedance of 50 in its passband. Assume that f∞ = 9.95 MHz. Plot its response in the frequency range 100 kHz to 100 MHz. 9.5. It is desired to design a maximally flat low-pass filter with at least 25 dB of attenuation at ω = 2ωc and −1.5 dB at its band edge. How many elements will be required for this filter? If a Chebyshev filter is used with 1.5 dB of ripple in its passband, find the number of circuit elements. 9.6. Design a low-pass Butterworth filter with a cutoff frequency of 150 MHz and an insertion loss of 50 dB at 400 MHz. It is to be used between a 75- load and a generator with its internal resistance as 75 . 9.7. Design a low-pass Chebyshev filter that exhibits ripples no more than 2 dB in its passband. The filter must pass all frequencies up to 50 MHz and attenuate the signal at 100 MHz by at least 15 dB. The load and the source resistance are of 50 each. 9.8. Reconsider Problem 9.7 to design a low-pass filter that exhibits the Chebyshev response with a 3-dB ripple in its passband. The filter must have 50 at both its input and output ports.

390

FILTER DESIGN

9.9. Design a high-pass Chebyshev filter with a passband ripple magnitude below 3 dB. It must pass all frequencies over 50 MHz. A minimum of 15 dB of insertion loss is required at 25 MHz. Assume that the load and source resistances are of 75 each. 9.10. Design a bandpass Chebyshev filter that exhibits no more than 2 dB of ripple in its passband. It must pass the signals that are in the frequency band 50 to 80 MHz with zero insertion loss. Assume that the load and source resistances are of 50 each. 9.11. It is desired to design a maximally flat low-pass filter with at least 40 dB of attenuation at f = 1.1 GHz and −3 dB at f = 1 GHz. How many elements will be required for this filter? If a Chebyshev filter is used with 3 dB of ripple in its passband, find the number of circuit elements. 9.12. Design a Butterworth filter with a cutoff frequency of 960 MHz and an insertion loss of 15 dB at 1.5 GHz. It is to be used between a 50- load and a generator with its internal resistance as 50 . 9.13. Design a low-pass Chebyshev filter that exhibits ripples of no more than 1 dB in its passband. The filter must pass all frequencies up to 960 MHz and attenuate the signal at 1.5 GHz by at least 15 dB. The load and the source resistance are of 50 each. 9.14. Reconsider Problem 9.13 to design a low-pass filter that exhibits the Chebyshev response with 2 dB of ripple in its passband. The filter must have 50 at both its input and output ports. 9.15. Design a high-pass Chebyshev filter with a passband ripple magnitude below 0.5 dB. It must pass all frequencies over 500 MHz. A minimum of 20 dB of insertion loss is required at 200 MHz. Assume that the load and source resistances are of 50 each. 9.16. Design a bandpass Chebyshev filter that exhibits no more than 0.1 dB of ripple in its passband. It must pass the signals that are in the frequency band 500 to 800 MHz with almost zero insertion loss and provide at least 15 dB of attenuation for signals below 300 MHz. Assume that the load and source resistances are of 75 each. 9.17. Design a high-pass filter that exhibits no more than 2.5 dB of ripple in its passband. It must pass all frequencies over 100 MHz. A minimum of 40 dB of insertion loss is required at 25 MHz. Assume that the load and source resistances are of 50 each. 9.18. Design a maximally flat bandstop filter with n = 5. It must stop the signals in the frequency range 50 to 80 MHz and pass the rest of the frequencies. Assume that the load and source resistances are of 50 each.

PROBLEMS

391

9.19. Design a fifth-order low-pass Butterworth filter with a cutoff frequency of 5 GHz. It is to be used between a 75- load and a generator with an internal resistance of 75 . 9.20. Design a third-order low-pass Chebyshev filter that exhibits ripples of no more than 1 dB in its passband. The filter must pass all frequencies up to 1 GHz. The load and the source resistance are of 50 each. 9.21. Design a fifth-order high-pass Chebyshev filter with passband ripple magnitude below 3 dB. It must pass all frequencies over 5 GHz. Assume that the load and source resistances are of 75 each.

10 SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

A signal-flow graph is a graphical means of portraying the relationship among the variables of a set of linear algebraic equations. S. J. Mason originally introduced it to represent the cause and effect of linear systems. Associated terms are defined in this chapter along with a procedure to draw a signal-flow graph for a given set of algebraic equations. Further, signal-flow graphs of microwave networks are obtained in terms of their S-parameters and associated reflection coefficients. The manipulation of signal-flow graphs is summarized to find the desired transfer functions. Finally, the relations for transducer power gain, available power gain, and operating power gain are formulated in this chapter. Consider a linear network that has N input and output ports. It is described by a set of linear algebraic equations: Vi =

N

Zij Ij

i = 1, 2, . . . , N

j =1

This says that the effect Vi at the ith port is a sum of gain times causes at its N ports. Hence, Vi represents the dependent variable (effect), and Ij are the independent variables (cause). Nodes or junction points of the signal-flow graph represent these variables. The nodes are connected by line segments called branches with an arrow on each directed toward the dependent node. The coefficient Zij is the gain of a branch that connects the ith dependent node with the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

392

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

393

j th independent node. A signal can be transmitted through a branch only in the direction of the arrow. The basic properties of a signal-flow graph can be summarized as follows: ž ž ž

ž ž ž

A signal-flow graph can be used only when the system is linear. A set of algebraic equations must be in the form of effects as functions of causes before its signal-flow graph can be drawn. A node is used to represent each variable. Normally, these are arranged from left to right, following a succession of inputs (causes) and outputs (effects) of the network. Nodes are connected together by branches with an arrow directed toward the dependent node. Signals travel along the branches only in the direction of the arrows. Signal Ik traveling along a branch that connects nodes Vi and Ik is multiplied by the branch gain Zik . The dependent node (effect) Vi is equal to the sum of the branch gain times the corresponding independent nodes (causes).

Example 10.1 The output b1 of a system is caused by two inputs a1 and a2 as represented by the equation b1 = S11 a1 + S12 a2 Find its signal-flow graph. SOLUTION There are two independent variables and one dependent variable in the equation. Locate these three nodes and connect nodes a1 and a2 with b1 , as shown in Figure 10.1. The arrows on two branches are directed toward effect b1 . Coefficients of input (cause) are shown as the gain of that branch. Example 10.2 Output b2 of a system is caused by two inputs a1 and a2 as represented by the equation b2 = S21 a1 + S22 a2 Find its signal-flow graph. a1

S11 a2

b1 S12

Figure 10.1

Signal-flow graph representation of Example 10.1.

394

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS a1

S21

a2 b2

Figure 10.2

S22

Signal-flow graph representation of Example 10.2.

SOLUTION There are again two independent variables and one dependent variable in the equation. Locate these three nodes and connect node a1 with b2 and a2 with b2 , as shown in Figure 10.2. The arrows on two branches are directed toward effect b2 . Coefficients of input (cause) are shown as the gain of that branch. Example 10.3 The input–output characteristics of a two-port network are given by the set of linear algebraic equations b1 = S11 a1 + S12 a2 b2 = S21 a1 + S22 a2 Find its signal-flow graph. SOLUTION There are two independent variables a1 and a2 , and two dependent variables b1 and b2 in this set of equations. Locate the four nodes and then connect nodes a1 and a2 with b1 . Similarly, connect a1 and a2 with b2 , as shown in Figure 10.3. Arrows on the branches are directed toward effects b1 and b2 . The coefficients of each input (cause) are shown as the gain of that branch. Example 10.4 The following set of linear algebraic equations represents the input–output relations of a multiport network. Find the corresponding signal-flow graph. S21 a1

b2 S11 S22 a2

b1 S12

Figure 10.3

Signal-flow graph representation of Example 10.3.

395

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

SOLUTION X1 = R1 +

1 X2 2+s

X2 = −4X1 + R2 − 7Y1 −

1 X2 s+4

s X2 +3 Y2 = 10X1 − sY1 Y1 =

s2

In the first equation, R1 and X2 represent the causes and X1 is the effect. Hence, the signal-flow graph representing this equation can be drawn as illustrated in Figure 10.4. Now consider the second equation. X1 is the independent variable in it. Further, X2 appears as cause as well as effect. This means that a branch must start and finish at the X2 node. Hence, when this equation is combined with the first, the signal-flow graph will look as illustrated in Figure 10.5. Next, we add to it the signal-flow graph of the third equation. It has Y1 as the effect and X2 as the cause. It is depicted in Figure 10.6. Finally, the last equation has Y2 as a dependent variable, and X1 and Y1 are two independent variables. A complete signal-flow graph representation is obtained after superimposing it as shown in Figure 10.7.

1 2+s

1 R1

X2 X1

Figure 10.4 Signal-flow graph representation of the first equation of Example 10.4.

−1 s+4 1 1 2+s

1 R1

R2

Y1

X1

X2

−7

−4

Figure 10.5 Signal-flow graph representation of the first two equations of Example 10.4.

396

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS −1 s+4

R2 1

1 2+s

1 R1

X2

X1

−4

Figure 10.6

Y1 −7

s s2 + 3

Signal-flow graph representation of the first three equations of Example 10.4.

−1 s+4 1 1 2+s

1 R1

R2

X1

X2

−7

−4

s s2 + 3

Y1

−s

10 Y2

Figure 10.7 Complete signal-flow graph representation of Example 10.4.

10.1 DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS Before we proceed with manipulation of signal-flow graphs, it will be useful to define a few remaining terms. Input and Output Nodes A node that has only outgoing branches is defined as an input node or source. Similarly, an output node or sink has only incoming branches. For example, R1 , R2 , and Y1 are the input nodes in the signal-flow graph shown in Figure 10.8. This corresponds to the first two equations of Example 10.4. There is no output node (exclude the dotted branches) in it because X1 and X2 have both outgoing as well as incoming branches. Nodes X1 and X2 in Figure 10.8 can be made the output nodes by adding an outgoing branch of

397

DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS

−1 s+4

R2

1 1 2+s

1 R1

Y1

X2

X1

−7 1

1 −4

X1

X2

Figure 10.8 Signal-flow graph with R1 , R2 , and Y1 as input nodes.

unity gain to each node. This is illustrated in Figure 10.8 by dotted branches. It is equivalent to adding X1 = X1 and X2 = X2 in the original set of equations. Thus, any nonoutput node can be made an output node in this way. However, this procedure cannot be used to convert these nodes to input nodes because that changes the equations. If an incoming branch of unity gain is added to node X1 , the corresponding equation is modified as follows: X1 = X1 + R1 +

1 X2 2+s

However, X1 can be made an input node by rearranging it as follows. The corresponding signal-flow graph is illustrated in Figure 10.9. It may be noted that R1 is now an output node: R1 = X1 −

1 X2 2+s

−1 s+4

R2 1

−

R1

1 2+s X2

1

Y1 −7

−4 X1

Figure 10.9 Signal-flow graph with R1 as an output, and X1 , R2 , and Y1 as the input nodes.

398

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Path A continuous succession of branches traversed in the same direction is called the path. It is known as a forward path if it starts at an input node and ends at an output node without hitting a node more than once. The product of branch gains along a path is defined as the path gain. For example, there are two forward paths between nodes X1 and R1 in Figure 10.9. One of these forward paths is just one branch connecting the two nodes with a path gain of 1. The other forward path is X1 to X2 to R1 . Its path gain is 4/(2 + s). Loop A loop is a path that originates and ends at the same node without encountering other nodes more than once along its traverse. When a branch originates and terminates at the same node, it is called a self-loop. The path gain of a loop is defined as the loop gain. Once the signal-flow graph is drawn, the ratio of an output to an input node (while other inputs, if there are more than one, are assumed to be zero) can be obtained by using rules of reduction. Alternatively, Mason’s rule may be used. However, the latter rule is prone to error if the signal-flow graph is too complex. The reduction rules are generally recommended for such cases and are given as follows: Rule 1 When there is only one incoming and one outgoing branch at a node (i.e., two branches are connected in series), it can be replaced by a direct branch with branch gain equal to the product of the two. This is illustrated in Figure 10.10. Rule 2 Two or more parallel paths connecting two nodes can be merged into a single path with a gain that is equal to the sum of the original path gains, as depicted in Figure 10.11. Rule 3 A self-loop of gain G at a node can be eliminated by multiplying its input branches by 1/(1 − G). This is shown graphically in Figure 10.12. Rule 4 A node that has one output and two or more input branches can be split in such a way that each node has just one input and one output branch, as shown in Figure 10.13. 5s

10s

2

X1

X1

X2

R1

X2

Figure 10.10 Graphical illustration of rule 1.

5s 5s + 3 X1

X2

X1

3

Figure 10.11 Graphical illustration of rule 2.

X2

399

DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS

2s 4

X2

X1

4 1 − 2s

X1

X2

Figure 10.12 Graphical illustration of rule 3.

X2

C2

X2

X′4 C4

C2 C4

X1 X4

C1

C1

X1

X5

X′′4

X5 C4

C3

C4

C3

X3

X′′′ 4

X3

Figure 10.13 Graphical illustration of rule 4.

X2

X4 C1

C2

X′′4

C4

C1

C2

X1

X′4

C4 X5

C3

X1

C1

X5

C3

C1 X3

X2

X3 X′′′ 4

Figure 10.14 Graphical illustration of rule 5.

Rule 5 This is similar to rule 4. A node that has one input and two or more output branches can be split in such a way that each node has just one input and one output branch. This is shown in Figure 10.14. Mason’s Gain Rule Ratio T of the effect (output) to that of the cause (input) can be found using Mason’s rule as follows: T (s) =

P1 1 + P2 2 + P3 3 + · · ·

where, Pi is the gain of the ith forward path, =1− L(1) + L(2) − L(3) + · · · 1 = 1 − L(1)(1) + L(2)(1) − L(3)(1) + · · ·

(10.1.1)

(10.1.2) (10.1.3)

400

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

2 = 1 − 3 = 1 −

L(1)(2) +

L(2)(2) −

L(3)(2) + · · ·

(10.1.4)

L(1)(3) + · · ·

(10.1.5)

.. . L(1) stands for the sum of all first-order loop gains, L(2) is the sum of all second-order loop gains, and so on. L(1)(1) denotes the sum of those first-order loop gains that do not touch path P1 at any node, L(2)(1) denotes the sum of those second-order loop gains that do not touch the path of P1 at any point, L(1)(2) denotes the sum of those first-order loops that do not touch path P2 at any point, and so on. First-order loop gain was defined earlier. Second-order loop gain is the product of two first-order loops that do not touch at any point. Similarly, third-order loop gain is the product of three first-order loops that do not touch at any point. Example 10.5 A signal-flow graph of a two-port network is given in Figure 10.15. Using Mason’s rule, find its transfer function Y/R. SOLUTION There are three forward paths from node R to node Y. Corresponding path gains are found as follows: s 3s 1 ×1 ×3×1= s +1 s +2 (s + 1)(s + 2) P2 = 1 × 6 × 1 = 6 P1 = 1 × 1 ×

and P3 = 1 × 1 ×

4 1 × (−4) × 1 = − s +1 s+1

Next, it has two loops. The loop gains are L1 = −

3 s+1

L2 = −

5s s+2

and

6 −4 1/(s + 1)

s/(s + 2)

R

Y 1

1

1 −3

Figure 10.15

3 −5

Signal-flow graph of Example 10.5.

1

SIGNAL-FLOW GRAPH REPRESENTATION OF A VOLTAGE SOURCE

401

Using Mason’s rule, we find that Y P1 + P2 (1 − L1 − L2 + L1 L2 ) + P3 (1 − L2 ) = R 1 − L1 − L2 + L1 L2

10.2 SIGNAL-FLOW GRAPH REPRESENTATION OF A VOLTAGE SOURCE Consider an ideal voltage source ES 0◦ in series with source impedance ZS , as shown in Figure 10.16. It is a single-port network with terminal voltage and current VS and IS , respectively. It is to be noted that the direction of current flow is assumed as entering the port, consistent with that of the two-port networks considered earlier. Further, the incident and reflected waves at this port are assumed to be aS and bS , respectively. The characteristic impedance at the port is assumed to be Z0 . Using the usual circuit analysis procedure, the total terminal voltage VS can be found as follows: VSin − VSref Z0 (10.2.1) where superscripts “in” and “ref” on VS and IS are used to indicate the corresponding incident and reflected quantities. This equation can be rearranged as follows: ZS ZS 1+ VSref = ES − 1 − VSin Z0 Z0 or Z0 Z0 − ZS in VSref = ES − V Z0 + ZS Z0 + ZS S VS = VSin + VSref = ES + ZS IS = ES + ZS (ISin + ISref ) = ES + ZS

IS

ZS

ES ∠0°

~

+

aS

VS

− bS

Figure 10.16 Incident and reflected waves at the output port of a voltage source.

402

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS 1

bS

1 bS

bG ΓS

aS aS

Figure 10.17

Dividing it by

√

1

Signal-flow graph representation for a voltage source.

2Z0 , we find that V ref Z0 Z0 − ZS VSin ES − √S = √ √ Z0 + ZS 2Z0 Z0 + ZS 2Z0 2Z0

Using (8.6.15) and (8.6.16), this can be written as follows: bS = bG + S aS

(10.2.2)

where V ref bS = √ S 2Z0 √ Z0 ES bG = √ 2(Z0 + ZS )

(10.2.3) (10.2.4)

V in aS = √ S 2Z0

(10.2.5)

and S =

ZS − Z0 ZS + Z0

(10.2.6)

From (10.2.2), the signal flow graph for a voltage source can be drawn as shown in Figure 10.17. 10.3 SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE Consider load impedance ZL as shown in Figure 10.18. It is a single-port device with port voltage and current VL and IL , respectively. The incident and reflected waves at the port are assumed to be aL and bL , respectively. Further, characteristic impedance at the port is Z0 . Following the usual circuit analysis rules, we find that VL = VLin + VLref = ZL IL = ZL (ILin + ILref ) = ZL

VLin − VLref Z0

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 403 IL aL

+ ZL

VL −

bL

Figure 10.18 Passive one-port circuit.

or 1+

ZL Z0

VLref =

ZL − 1 VLin Z0

(10.3.1)

or VLref = After dividing by

√

ZL − Z0 in V = L VLin ZL + Z0 L

(10.3.2)

2Z0 , we use (8.6.15) and (8.6.16) to find that bL = L aL

(10.3.3)

where V ref bL = √ L 2Z0

(10.3.4)

V in aL = √ L 2Z0

(10.3.5)

and L =

ZL − Z0 ZL + Z0

(10.3.6)

A signal-flow graph can be drawn on the basis of (10.3.3), as shown in Figure 10.19.

1 aL

aL ΓL

bL

bL 1

Figure 10.19 Signal-flow graph of a one-port passive device.

404

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Γin

Two-port network

ZL

Figure 10.20 Two-port network with termination.

Example 10.6 Impedance ZL terminates port 2 (the output) of a two-port network as shown in Figure 10.20. Draw the signal-flow graph and determine the reflection coefficient at its input port using Mason’s rule. SOLUTION As shown in Figure 10.21, combining the signal-flow graphs of a passive load and that of a two-port network obtained in Example 10.3, we can get the representation for this network. Its input reflection coefficient is given by the ratio of b1 to a1 . For Mason’s rule we find that there are two forward paths, a1 → b1 and a1 → b2 → aL → bL → a2 → b1 . There is one loop, b2 → aL → bL → a2 → b2 , which does not touch the former path. The corresponding path and loop gains are P1 = S11 P2 = S21 × 1 × L × 1 × S12 = L S21 S12 and L1 = 1 × L × 1 × S22 = L S22 Hence, b1 P1 (1 − L1 ) + P2 S11 (1 − L S22 ) + L S21 S12 = = a1 1 − L1 1 − L S22 L S21 S12 = S11 + 1 − L S22

in =

a1

Two-port network S21 b2

S11

Load 1

aL

S22

b1

S12

a2

ΓL

1

bL

Figure 10.21 Signal-flow graph representation of the network shown in Figure 10.20.

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 405 ZS Γout VS ∠0° ~

Two-port network

ZL

Figure 10.22 Two-port network with the source and termination.

Example 10.7 A voltage source is connected at the input port of a two-port network and the load impedance ZL terminates its output, as shown in Figure 10.22. Draw its signal-flow graph and find the output reflection coefficient out . SOLUTION A signal-flow graph of this circuit can be drawn by combining the results of Example 10.6 with those of a voltage source representation obtained in Section 10.2. This is illustrated in Figure 10.23. The output reflection coefficient is defined as the ratio of b2 to a2 with load disconnected, and the source impedance ZS terminates port 1 (the input). Hence, there are two forward paths from a2 to b2 , a2 → b2 , and a2 → b1 → aS → bS → a1 → b2 . There is only one loop (because the load is disconnected), b1 → aS → bS → a1 → b1 . The path and loop gains are found as follows: P1 = S22 P2 = S12 × 1 × S × 1 × S21 = S S12 S21 and L1 = 1 × S × 1 × S11 = S S11 Therefore, from Mason’s rule, we have b2 P1 (1 − L1 ) + P2 S22 (1 − S11 S ) + S21 S12 S = = a2 1 − L1 1 − S11 S S21 S12 S = S22 + 1 − S11 S

out =

Voltage source

Two-port network bS

1

a1

1

S21

Load b2

aL

1

bG ΓS

S11

aS

1

ΓL

S22

b1

S12

a2

1

bL

Figure 10.23 Signal-flow graph representation of the network shown in Figure 10.22.

406

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS Voltage source 1

bS

Load aL

1

bG ΓS

aS

ΓL

1

bL

Figure 10.24 Signal-flow graph for Example 10.8.

Example 10.8 The signal-flow graph shown in Figure 10.24 represents a voltage source that is terminated by a passive load. Analyze the power transfer characteristics of this circuit and establish the conditions for maximum power transfer. SOLUTION Using the defining equations of ai and bi , we can develop the following relations: power output of the source = |bS |2 power reflected back into the source = |aS |2 Hence, power delivered by the source, Pd , is Pd = |bS |2 − |aS |2 Similarly, we can write the following relations for power at the load: power incident on the load = |aL |2 power reflected from the load = |bL |2 = |L |2 |aL |2 where L is the load reflection coefficient. Hence, power absorbed by the load, PL , is given by PL = |aL |2 − |bL |2 = |aL |2 (1 − |L |2 ) = |bS |2 (1 − |L |2 ) Since bS = bG + S aS = bG + S bL = bG + S L aL = bG + S L bS we find that bS =

bG 1 − S L

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 407

and

1 1 aS = (bS − bG ) = S S

bG − bG 1 − S L

=

L bG 1 − S L

Hence, 2 L bG 2 b G − Pd = |bS | − |aS | = 1 − 1 − S L S L 2 bG (1 − |L |2 ) = |bS |2 (1 − |L |2 ) = PL = 1 − S L 2

2

This simply says that the power delivered by the source is equal to the power absorbed by the load. We can use this equation to establish the load condition under which the source delivers maximum power. For that, we consider ways to maximize. 2 bG (1 − |L |2 ) PL = 1− S

L

For PL to be a maximum, its denominator 1 − S L is minimized. Let us analyze this term more carefully using the graphical method. It says that there is a phasor S L at a distance of unity from the origin that rotates with a change in L (S is assumed constant), as illustrated in Figure 10.25. Hence, the denominator has an extreme value whenever S L is a pure real number. If this number is positive real, the denominator is minimum. On the other hand, the denominator is maximized when S L is a negative real number. For S L to be a positive real number, the load reflection coefficient L must be the complex conjugate of the source reflection coefficient S . In other words, L = S∗ ⇒

ZL − Z0 = ZL + Z0

ZS − Z0 ZS + Z0

∗

1 ΓSΓL

1 − ΓSΓL

Figure 10.25 Graphical representation of (1 − S L ).

408

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

For a real Z0 it reduces to

ZL = ZS∗

The value of this maximum power is PL =

|bG |2 = Pavs (1 − |S |2 )

where Pavs is the maximum power available from the source. It is a well-known maximum power transfer theorem of electrical circuit theory. Example 10.9 Draw the signal-flow graph of the three-port network shown in Figure 10.26. Find the transfer characteristics b3 /bS using the graph. SOLUTION For a three-port network, the scattering equations are given as follows: b1 = S11 a1 + S12 a2 + S13 a3 b2 = S21 a1 + S22 a2 + S23 a3 b3 = S31 a1 + S32 a2 + S33 a3 For the voltage source connected at port 1, we have bS = bG + S aS Since the wave coming out from the source is incident on port 1, bS is equal to a1 . Therefore, we can write a1 = bG + S b1 For the load ZL connected at port 2, bL = L aL

ZD ZS

VS ∠0° ~

Port 1

Port 3

Three-port network

Port 2

Figure 10.26 Three-port network of Example 10.9.

ZL

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 409

Again, the wave emerging from port 2 of the network is incident on the load. Similarly, the wave reflected back from the load is incident on port 2 of the network. Hence, b2 = aL and bL = a2 Therefore, the relation above at the load ZL can be modified as follows: a2 = L b2 Following a similar procedure for the load ZD connected at port 3, we find that bD = D aD ⇒ a3 = D b3 Using these equations, a signal-flow graph can be drawn for this circuit as shown in Figure 10.27. For b3 /bs , there are two forward paths and eight loops in this flow graph. Path gains Pi and loop gains Li are found as follows: P1 = S31 P2 = S21 S32 L ΓD

b3

a3 S33

S31

S23 S13

1

bS

S32

a1

b2 1

aL

bG 1 ΓS

aS

Figure 10.27

S21

S11

1

ΓL

S22

b1

S12

a2

1

bL

Signal-flow graph of the network shown in Figure 10.26.

410

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

L1 = S11 S L2 = S22 L L3 = S33 D L4 = S21 S12 L S L5 = S31 S23 S12 D L S L6 = S13 S21 S32 D L S L7 = S23 S32 D L and L8 = S13 S31 D S Therefore, various terms of Mason’s rule can be evaluated as follows: 1 = 1 − S22 L 2 = 1 L(1) = L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8 L(2) = L1 L2 + L1 L3 + L2 L3 + L3 L4 + L1 L7 + L2 L8 L(3) = L1 L2 L3 Using Mason’s rule we can find that b3 P1 1 + P2 2 = bS 1 − L(1) + L(2) − L(3) Note that this signal-flow graph is too complex, and therefore one can easily miss a few loops. In cases like this, it is prudent to simplify the signal-flow graph using the five rules mentioned earlier.

10.4 POWER GAIN EQUATIONS We have seen in the preceding section that maximum power is transferred when load reflection coefficient L is the conjugate of the source reflection coefficient S . It can be generalized for any port of the network. In the case of an amplifier, maximum power will be applied to its input port if its input reflection coefficient in is the complex conjugate of the source reflection coefficient S . If this is not the case, the input will be less than the maximum power available from the source. Similarly, maximum amplified power will be transferred to the load only if load reflection coefficient L is the complex conjugate of output reflection coefficient out . Part of the power available at the output of the amplifier will be

411

POWER GAIN EQUATIONS

reflected back if there is a mismatch. Therefore, the power gain of an amplifier can be defined at least three different ways as follows: PL power delivered to the load = Pavs power available from the source

transducer power gain, GT = operating power gain, GP =

PL power delivered to the load = Pin power input to the network

available power gain, GA =

PAVN power available from the network = Pavs power available from the source

and

where Pavs is the power available from the source, Pin the power input to the network, PAVN the power available from the network, and PL the power delivered to the load. Therefore, the transducer power gain will be equal to that of operating power gain if the input reflection coefficient in is the complex conjugate of the source reflection coefficient S . That is, Pavs = Pin |in =S∗ Similarly, the available power gain will be equal to the transducer power gain if the following condition holds at the output port: ∗ PAVN = PL |L =out

In this section we formulate these power gain relations in terms of scattering parameters and reflection coefficients of the circuit. To facilitate the formulation, we reproduce in Figure 10.28 the signal-flow graph obtained in Example 10.7.

Amplifier circuit (two-port)

Voltage source 1

bS

a1

1

S21

Load 1

b2

aL

bG S11

ΓS

aS

Pavs

1

Pin

ΓL

S22

b1

S12

a2

PAVN

1

bL

PL

Figure 10.28 Signal-flow graph for an amplifier circuit with voltage source and terminating load.

412

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Transducer Power Gain To develop the relation for transducer power gain, we first determine the power delivered to load PL and the power available from the source Pavs : PL = |aL |2 − |bL |2 = |aL |2 (1 − |L |2 ) = |b2 |2 (1 − |L |2 )

(10.4.1)

We derived an expression for Pavs in Example 10.7. That is rewritten here: Pavs = |bS |2 − |aS |2 =

|bG |2 1 − |S |2

(10.4.2)

Therefore, the transducer power gain is GT =

|b2 |2 (1 − |L |2 )(1 − |S |2 ) |bG |2

(10.4.3)

We now need to find an expression for b2 /bG . Mason’s rule can be used to formulate it. There is only one forward path in this case. However, there are three loops. The path gain P1 and loop gains Li are P1 = S21 L1 = S11 S L2 = S22 L and L3 = S12 S21 S L Therefore, P1 b2 = bG 1 − (L1 + L2 + L3 ) + L1 L2 S21 = 1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L

(10.4.4)

Substituting (10.4.4) into (10.4.3), we get GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L |2

It can be simplified as follows: GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S11 S )(1 − S22 L ) − S21 S12 L S |2

(10.4.5)

413

POWER GAIN EQUATIONS

or GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S22 L )[1 − S11 S − S21 S12 L S /(1 − S22 L )]|2

GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S22 L )(1 − in S )|2

or

=

1 − |S |2 1 − |L |2 2 |S | 21 |1 − in S |2 |1 − S22 L |2

where in = S11 +

S21 S12 L 1 − S22 L

(10.4.6)

(10.4.7)

This is the input reflection coefficient formulated in Example 10.6. Alternatively, (10.4.5) can be written as GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) 1 − |S |2 1 − |L |2 = |S21 |2 2 2 |(1 − S11 S )(1 − out L )| |1 − S11 S | |1 − out L |2 (10.4.8)

where out = S22 +

S21 S12 S 1 − S11 S

(10.4.9)

This is the output reflection coefficient formulated in Example 10.7. Special Case Consider the case when the two-port network is unilateral. This means that there is an output signal at port 2 when a source is connected at port 1. However, a source (cause) connected at port 2 does not show its response (effect) at port 1. In terms of scattering parameters of the two-port, S12 is equal to zero in this case, and therefore, the input and output reflection coefficients in (10.4.7) and (10.4.9) simplify as follows: in = S11 and

out = S22

The transducer power gain GT as given by (10.4.6) or (10.4.8) simplifies to GT |S12 =0 = GT U =

1 − |S |2 1 − |L |2 2 |S | = GS G0 GL 21 |1 − S11 S |2 |1 − S22 L |2

(10.4.10)

where GT U may be called the unilateral transducer power gain, while GS =

1 − |S |2 |1 − S11 S |2

G0 = |S21 |2

(10.4.11) (10.4.12)

414

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

and GL =

1 − |L |2 |1 − S22 L |2

(10.4.13)

Equation (10.4.10) indicates that maximum GT U is obtained when the input and output ports are conjugate matched. Hence, GT Umax = GSmax G0 GLmax

(10.4.14)

GSmax = GS |S =S11∗

(10.4.15)

GLmax = GL |L =S22∗

(10.4.16)

where

and

Operating Power Gain To find the operating power gain, we require the power delivered to the load and power input to the circuit. We evaluated the power delivered to the load in (10.4.1). The power input to the circuit is found as follows: Pin = |a1 |2 − |b1 |2 = |a1 |2 (1 − |in |2 ) Hence, GP =

PL |b2 |2 (1 − |L |2 ) = Pin |a1 |2 (1 − |in |2 )

(10.4.17)

Now we evaluate b2 /a1 using Mason’s rule. With the source disconnected, we find that there is only one forward path and one loop with gains S21 and S22 L , respectively. Hence, b2 S21 = (10.4.18) a1 1 − S22 L Substituting (10.4.18) into (10.4.17), we find that GP =

1 1 − |L |2 |S21 |2 2 1 − |in | |1 − S22 L |2

(10.4.19)

Available Power Gain Expressions for maximum powers available from the network and the source are needed to determine the available power gain. Maximum power available from the network PAVN is delivered to the load if the load reflection coefficient is a complex conjugate of the output. Using (10.4.1), we find that 2 2 2 2 ∗ = |b | (1 − | | ) = |b | (1 − | PAVN = PL |L =out 2 L 2 out | )

(10.4.20)

415

POWER GAIN EQUATIONS

The power available from the source is given by (10.4.2). Hence, GA =

PAVN |b2 |2 = (1 − |out |2 )(1 − |S |2 ) Pavs |bG |2

(10.4.21)

Next, using Mason’s rule in the signal-flow graph of Figure 10.28, we find that S21 b2 = bG 1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L S21 = (1 − S11 S )(1 − out L ) and

S21 b2 = bG L =out (1 − S11 S )(1 − |out |2 ) ∗

(10.4.22)

Now, substituting (10.4.22) into (10.4.21), we get GA =

1 − |S |2 1 |S21 |2 2 |1 − S11 S | 1 − |out |2

(10.4.23)

Example 10.10 The scattering parameters of a microwave amplifier are found at 800 MHz as follows (Z0 = 50 ): S11 = 0.45 150◦ , S12 = 0.01 − 10◦ , S21 = 2.05 10◦ , and S22 = 0.4 − 150◦ . The source and load impedances are 20 and 30 , respectively. Determine the transducer power gain, operating power gain, and available power gain. SOLUTION First we determine the source reflection coefficient S and the load reflection coefficient L as follows: S =

ZS − Z0 20 − 50 30 = =− = −0.429 ZS + Z0 20 + 50 70

L =

ZL − Z0 30 − 50 20 = =− = −0.25 ZL + Z0 30 + 50 80

and

Now the input reflection coefficient in and the output reflection coefficient out can be calculated from (10.4.7) and (10.4.9), respectively: in = S11 + or

S21 S12 L (0.01 − 10◦ )(2.05 10◦ )(−0.25) ◦ = 0.45 150 + 1 − S22 L 1 − (0.4 − 150◦ )(−0.25) ◦

in = −0.3953 + j 0.2253 = 0.455 150.32

416

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

and S21 S12 S = 0.4 1 − S11 S

out = S22 +

◦

− 150 +

(0.01 − 10◦ )(2.05 10◦ )(−0.429) 1 − (0.45 150◦ )(−0.429)

or out = −0.3568 − j 0.1988 = 0.4084

◦

− 150.87

The transducer power gain GT can now be calculated from (10.4.6) or (10.4.8) as follows: |S21 |2 (1 − |L |2 )(1 − |S |2 ) |S21 |2 (1 − |L |2 )(1 − |S |2 ) = |(1 − S22 L )(1 − in S )|2 |(1 − S11 S )(1 − out L )|2

GT = Hence,

1 − (0.429)2 1 − (0.252 ) 2 (2.05) |1 − (0.45 150◦ )(−0.429)|2 |1 − (0.4084 − 150.87◦ )(−0.25)|2

GT =

= 5.4872 Similarly, from (10.4.19), GP =

1 1 − |L |2 2 |S | 21 1 − |in |2 |1 − S22 L |2

Therefore, GP =

1 1 − (0.252 ) 2 ) = 5.9374 (2.05 1 − (0.4552 ) |1 − (0.4 − 150◦ )(−0.25)|2

The available power gain is calculated from (10.4.23) as follows: GA =

1 − |S |2 1 |S21 |2 2 |1 − S11 S | 1 − |out |2

Hence, GA =

1 − (0.429)2 1 (2.052 ) = 5.8552 |1 − (0.45 150◦ )(−0.429)|2 1 − (0.4084)2

The three power gains can be expressed in decibels as follows: GT (dB) = 10 log(5.4872) ≈ 7.4 dB GP (dB) = 10 log(5.9374) ≈ 7.7 dB and GA (dB) = 10 log(5.8552) ≈ 7.7 dB

417

SUGGESTED READING

Example 10.11 The scattering parameters of a microwave amplifier are found at 2 GHz as follows (Z0 = 50 ): S11 = 0.97 − 43◦ , S12 = 0.0, S21 = 3.39 140◦ , and S22 = 0.63 − 32◦ . The source and load reflection coefficients are 0.97 43◦ and 0.63 32◦ , respectively. Determine the transducer power gain, operating power gain, and available power gain. SOLUTION Since S12 is zero, this amplifier is unilateral. Therefore, its input and output reflection coefficients from (10.4.7) and (10.4.9) may be found as follows: ◦

input reflection coefficient in = S11 = 0.97

− 43

output reflection coefficient out = S22 = 0.63

− 32

◦

On comparing the input reflection coefficient with that of the source, we find that one is the complex conjugate of the other. Hence, the source is matched with the input port, and therefore the power available from the source is equal to the power input. Similarly, we find that the load reflection coefficient is the conjugate of the output reflection coefficient. Hence, power delivered to the load will be equal to power available from the network at port 2. As may be verified from (10.4.10), (10.4.19), and (10.4.23), these conditions make all three power gains equal. From (10.4.10), GT U =

1 − |L |2 1 − |S |2 1 − (0.97)2 2 |S | = (3.39)2 21 |1 − S11 S |2 |1 − S22 L |2 |1 − (0.97)2 |2 ×

1 − (0.63)2 = 322.42 |1 − (0.63)2 |2

From (10.4.19), GP =

2 1 1 1 − |L |2 2 2 1 − (0.63) |S | = (3.39) 21 1 − |in |2 |1 − S22 L |2 1 − (0.97)2 |1 − (0.63)2 |2

= 322.42 From (10.4.23), GA =

1 − |S |2 1 − (0.97)2 1 1 |S21 |2 = (3.39)2 2 2 2 |1 − S11 S | 1 − |out | |1 − (0.97) | 1 − (0.63)2

= 322.42

SUGGESTED READING Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Davis, W. A., Microwave Semiconductor Circuit Design. New York: Van Nostrand Reinhold, 1984.

418

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Gonzalez, G., Microwave Transistor Amplifiers. Upper Saddle River, NJ: Prentice Hall, 1997. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Wolff, E. A., and R. Kaul, Microwave Engineering and Systems Applications. New York: Wiley, 1988.

PROBLEMS 10.1. Use Mason’s rule to find Y (s)/R(s) for the signal-flow graph shown in Figure P10.1. 3

1

2

1/S

1

1/S

1 Y(s)

R(s)

−4 −5

Figure P10.1

10.2. Use Mason’s gain rule to find Y5 (s)/Y1 (s) and Y2 (s)/Y1 (s) for the system shown in Figure P10.2. G4

1

Y2

G1

Y3

G2

Y4

G3

Y6

1 Y5

Y1 −H1

−H2

−H3

Figure P10.2

419

PROBLEMS

10.3. Use Mason’s gain rule to find Y2 (s)/R3 (s) for the system shown in Figure P10.3. R2(s) R3(s)

1

1/S

Y3

Y4

6

Y6

1/S

1 Y1

R1(s) −3

−2 Y2(s) −4

Figure P10.3

10.4. Use Mason’s gain rule to find Y (s)/R(s) for the signal-flow graph shown in Figure P10.4. R(s)

1

4 3

2

1/(S + 1)

1 1/(S + 1) 6 7

5

Y(s) 1

Figure P10.4

10.5. Draw the signal-flow graph for the circuit shown in Figure P10.5. Use Mason’s gain rule to determine the power ratios P2 /P1 and P3 /P1 . Port 3 P3 Port 1

[S]

P1

Port 2

Figure P10.5

0 [S] = S21 0

S12 0 S32

0 S23 0

P2

420

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

in is the reflection coefficient at port 1, and 2 and 3 are the reflection coefficients at ports 2 and 3, respectively. 10.6. A voltage source has an internal impedance of 100 , and an electromotive force of 3 V. Using Z0 = 50 , find the reflection coefficient and scattering variables a and b for the source. 10.7. A 10-mV voltage source has an internal impedance of 100 + j 50 . Assuming that the characteristic impedance Z0 is 50 , find the reflection coefficient and scattering variables a and b for the source. Draw the signal-flow graph. 10.8. A voltage source of 1 0◦ V has an internal resistance of 80 . It is connected to a load of 30 + j 40 . Assuming that the characteristic impedance is 50 , find the scattering variables and draw the signal-flow graph. 10.9. The scattering parameters of a two-port network are S11 = 0.26 − j 0.16, S12 = S21 = 0.42, and S22 = 0.36 − j 0.57. (a) Determine the input reflection coefficient and transducer power loss for ZS = ZL = Z0 . (b) Do the same for ZS = Z0 and ZL = 3Z0 . 10.10. Two two-port networks are connected in cascade. The scattering parameters of the first two-port network are S11 = 0.28, S12 = S21 = 0.56, and S22 = 0.28 − j 0.56. For the second network, S11 = S22 = −j 0.54 and S12 = S21 = 0.84. Determine the overall scattering and transmission parameters. 10.11. Draw the flow graph representation for the two cascaded circuits of Problem 10.10. Using Mason’s rule, determine the following: (a) the input reflection coefficient when the second network is match-terminated and (b) the forward transmission coefficient of the overall network. 10.12. A GaAs MESFET has the following S-parameters measured at 8 GHz with a 50- reference: S11 = 0.26 − 55◦ , S12 = 0.08 80◦ , S21 = 2.14 65◦ , and S22 = 0.82 − 30◦ . For ZL = 40 + j 160 and S = 0.31 − 76.87◦ , determine the operating power gain of the circuit. 10.13. The S-parameters of a certain microwave transistor measured at 3 GHz with a 50- reference are S11 = 0.406 − 100◦ , S12 = 0, S21 = 5.0 50◦ , and S22 = 0.9 − 60◦ . Calculate the maximum unilateral power gain. 10.14. (a) Find the value of the source impedance that results in maximum power delivered to the load in the circuit shown in Figure P10.14. Evaluate the maximum power delivered to the load.

421

PROBLEMS Zs

10 V ∠0°

Z0 = 50 Ω

ZL = 50 + j50 Ω

λ /4

Figure P10.14

(b) Using the value of Zs from part (a), find the Th´evenin equivalent circuit at the load end and evaluate the power delivered to the load.

11 TRANSISTOR AMPLIFIER DESIGN

Amplifiers are among the basic building blocks of an electronic system. Although vacuum-tube devices are still used in high-power microwave circuits, transistors—silicon bipolar junction devices, GaAs MESFETs, heterojunction bipolar transistors (HBTs), and high-electron mobility transistors (HEMTs)—are common in many radio-frequency and microwave designs. The chapter begins with the stability considerations for a two-port network and the formulation of relevant conditions in terms of its scattering parameters. Expressions for input and output stability circles are presented next to facilitate the design of amplifier circuits. Design procedures for various small-signal single-stage amplifiers are discussed for unilateral as well as bilateral transistors. Noise figure considerations in amplifier design are discussed in the following section. An overview of broadband amplifiers is included. Small-signal equivalent circuits and biasing mechanisms for various transistors are also summarized in subsequent sections.

11.1 STABILITY CONSIDERATIONS Consider a two-port network that is terminated by load ZL as shown in Figure 11.1. A voltage source VS with internal impedance ZS is connected at its input port. Reflection coefficients at its input and output ports are in and out , respectively. The source reflection coefficient is S and the load reflection coefficient is L . Expressions for input and output reflection coefficients were formulated in Examples 10.6 and 10.7. For this two-port to be unconditionally stable at a given frequency, the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

422

423

STABILITY CONSIDERATIONS ZS

VS ∠0°

~

ZL

Two-port network

ΓS Zin, Γin

Zout, Γout ΓL

Figure 11.1 Two-port network with voltage source at its input and a load terminating the output port.

following inequalities must hold:

and

|S | < 1

(11.1.1)

|L | < 1 S21 S12 L 0 or 1 − |S11 |2 − |S22 |2 + ||2 > 2|S12 S21 | or

1 − |S11 |2 − |S22 |2 + ||2 >1 2|S12 S21 |

Therefore, k=

1 − |S11 |2 − |S22 |2 + ||2 >1 2|S12 S21 |

(11.1.18)

426

TRANSISTOR AMPLIFIER DESIGN

If the S-parameters of a transistor satisfy conditions (11.1.16) and (11.1.18), it is stable for any passive load and generator impedance. In other words, this transistor is unconditionally stable. On the other hand, it may be conditionally stable (stable for limited values of load or source impedance) if one or both of these conditions are violated. It means that the transistor can provide stable operation for a restricted range of S and L . A simple procedure to find these stable regions is to test inequalities (11.1.3) and (11.1.4) for particular load and source impedances. An alternative graphical approach is to find the circles of instability for load and generator reflection coefficients on a Smith chart. The latter approach is presented below. From the expression of input reflection coefficient (10.4.7), we find that in = S11 +

S21 S12 L ⇒ in (1 − S22 L ) = S11 (1 − S22 L ) + S21 S12 L 1 − S22 L

or in = S11 − L (S11 S22 − S12 S21 − in S22 ) ⇒ L =

S11 − in − in S22

or L =

S11 − in S22 1 S11 S22 − in S22 − S12 S21 + S12 S21 = − in S22 S22 S22 − in S22

L =

1 S22

or

1+

S12 S21 − in S22

=

1 S22

+

S12 S21 1 − in −1 S22

(11.1.19)

As before, 1 − in S22 −1 represents a circle on the complex plane. It is centered at 1 with radius |in S22 −1 |; the reciprocal of this expression is another circle with center at 1 1 1 1 + = −1 −1 2 1 + | S22 | 1 − | S22 | 1 − |−1 S22 |2 and radius 1 2

1 1 − 1 − |−1 S22 | 1 + |−1 S22 |

=

|−1 S22 | 1 − |−1 S22 |2

Since |in | < 1, the region of stability will include all points on the Smith chart outside this circle. From (11.1.19), the center of the load impedance circle, CL , is CL =

1 S22

or 1 CL = S22

+

S12 S21 1 − |−1 S22 |2

S12 S21 ∗ 1+ ||2 − |S22 |2

=

=

1 S22

+

S12 S21 ||2 ||2 − |S22 |2

1 ||2 − |S22 |2 + S12 S21 ∗ S22 ||2 − |S22 |2

STABILITY CONSIDERATIONS

or CL =

427

1 ∗ ( + S12 S21 ) − |S22 |2 1 ∗ S11 S22 − |S22 |2 = S22 ||2 − |S22 |2 S22 ||2 − |S22 |2

Therefore, CL =

∗ ∗ ∗ ∗ S11 − S22 (S22 − S11 ) = ||2 − |S22 |2 |S22 |2 − ||2

(11.1.20)

Its radius, rL , is given by 1 S12 S21 |−1 S22 | S12 S21 = rL = |S22 | 1 − |−1 S22 |2 ||2 − |S22 |2

(11.1.21)

As explained following (11.1.19), this circle represents the locus of points over which the input reflection coefficient in is equal to unity. On one side of this circle, the input reflection coefficient is less than unity (stable region); on its other side it exceeds 1 (unstable region). When load reflection coefficient L is zero (i.e., a matched termination is used), in is equal to S11 . Hence, the center of the Smith chart (reflection coefficient equal to zero) represents a stable point if |S11 | is less than unity. On the other hand, it represents unstable impedance for |S11 | greater than unity. If L = 0 is located outside the stability circle and is found stable, all outside points are stable. Similarly, if L = 0 is inside the stability circle and is found stable, all enclosed points are stable. If L = 0 is unstable, all points on that side of the stability circle are unstable. Similarly, the locus of S can be derived from (11.1.4), with its center CS and its radius rS given as CS = and rS =

∗ ∗ ∗ ∗ S22 − S11 (S11 − S22 ) = 2 2 2 || − |S11 | |S11 | − ||2

1 S12 S21 |−1 S11 | S12 S21 = |S11 | 1 − |−1 S11 |2 ||2 − |S11 |2

(11.1.22)

(11.1.23)

This circle represents the locus of points over which output reflection coefficient out is equal to unity. On one side of this circle, the output reflection coefficient is less than unity (stable region), whereas on its other side it exceeds 1 (unstable region). When the source reflection coefficient S is zero, out is equal to S22 . Hence, the center of the Smith chart (reflection coefficient equal to zero) represents a stable point if |S22 | is less than unity. On the other hand, it represents an unstable impedance point for |S22 | greater than unity. If S = 0 is located outside the stability circle and is found stable, all outside source-impedance points are stable. Similarly, if S = 0 is inside the stability circle and is found stable, all enclosed points are stable. If S = 0 is unstable, all points on that side of the stability circle are unstable.

428

TRANSISTOR AMPLIFIER DESIGN

Example 11.1 The S-parameters of a properly biased transistor are found at ◦ 2 GHz as follows (50- reference impedance): S11 = 0.894 −60.6 , S12 = ◦ ◦ ◦ 0.02 62.4 , S21 = 3.122 123.6 , and S22 = 0.781 −27.6 . Determine its stability and plot the stability circles if the transistor is potentially unstable (see Figure 11.3). SOLUTION From (11.1.16) and (11.1.18), we get || = |S11 S22 − S12 S21 | = 0.6964 and

1 + ||2 − |S11 |2 − |S22 |2 = 0.6071 2|S12 S21 |

k=

Since one of the conditions for stability failed above, this transistor is potentially unstable.

0 12

110

90

100

rS

80

CL

70 CS 60 50

0

rL

10

170

20

160

15

30

0

14

40

0

13

10

5

2

1

0.5

−10

70 0 −1

0.2

0

180

0

50 −1

−16

−20 0

−1

−4

40

−3 0

−5

0

30

−6

0

−70

20

−80

0

−90

−100 −11

−1

−1

Figure 11.3

Input and output stability circles for Example 11.1.

AMPLIFIER DESIGN FOR MAXIMUM GAIN

429

Using (11.1.20) to (11.1.23), we can determine the stability circles as follows. For the output stability circle CL = 1.36 46.7◦ and rL = 0.5. Since |S11 | is 0.894, L = 0 represents a stable load point. Further, this point is located outside the stability circle, and therefore all points outside this circle are stable. For the input stability circle CS = 1.13 68.5◦ and rS = 0.2. Since |S22 | is 0.781, S = 0 represents a stable source impedance point. Further, this point is located outside the stability circle, and therefore all points outside this circle are stable. For the output stability circle, draw a radial line at 46.7◦ . With the radius of the Smith chart as unity, locate the center at 1.36 on this line. It can be done as follows. Measure the radius of the Smith chart using a ruler scale. Supposing that it is d millimeters, the location of the stability circle is then at 1.36d millimeters away on this radial line. Similarly, the radius of the stability circle is 0.5d millimeters. The load impedance must lie outside this circle for the circuit to be stable. Following a similar procedure, the input stability circle is drawn with its radius as 0.2d millimeters and center at 1.13d millimeters on the radial line at 68.5◦ . To have a stable design, the source impedance must lie outside this circle.

11.2 AMPLIFIER DESIGN FOR MAXIMUM GAIN In this section we first consider the design of an amplifier that uses a unilateral transistor (S12 = 0) and has maximum possible gain. A design procedure using a bilateral transistor (S12 = 0) is developed next that requires simultaneous conjugate matching at its two ports. Unilateral Case When S12 is zero, the input reflection coefficient in reduces to S11 and the output reflection coefficient out simplifies to S22 . To obtain maximum gain, the ∗ ∗ and S22 , respectively. source and load reflection coefficients must be equal to S11 Further, the stability conditions for a unilateral transistor simplify to |S11 | < 1 and |S22 | < 1. Example 11.2 The S-parameters of a properly biased BJT are found at 1 GHz ◦ ◦ as follows (with Z0 = 50 ): S11 = 0.60 −155 , S22 = 0.48 −20 , S12 = 0, ◦ and S21 = 6 180 . Determine the maximum gain possible with this transistor and design an RF circuit that can provide this gain. SOLUTION 1. Stability check: k=

1 − |S11 |2 − |S22 |2 + ||2 =∞ 2|S12 S21 |

430

TRANSISTOR AMPLIFIER DESIGN

because S12 = 0 and || = |S11 S22 − S12 S21 | = |S11 S22 | = 0.2909 Since both of the conditions are satisfied, the transistor is unconditionally stable. 2. The maximum possible power gain of the transistor is found as GTU =

2 1 − |s |2 2 1 − |L | |S | 21 |1 − S11 s |2 |1 − S22 L |2

and ∗ 2 ∗ 2 1 − |S11 | 2 1 − |S22 | |S | 21 |1 − |S11 |2 |2 |1 − |S22 |2 |2 2

1 1 = = 73.9257 6 1 − 0.6062 1 − 0.482

GT Umax =

or GTUmax = 10 log10 (73.9257) dB = 18.688 dB 3. For the maximum unilateral power gain, ◦

∗ S = S11 = 0.606 155

◦

∗ and L = S22 = 0.48 20

The component values are determined from the Smith chart as illustrated in Figure 11.4. Corresponding to the reflection coefficient’s magnitude 0.606, the VSWR is 4.08. Hence, point D in Figure 11.4 represents the source impedance ZS . Similarly, point B can be identified as the load impedance ZL . Alternatively, the normalized impedance can be computed from the reflection coefficient. Then, points D and B can be located on the Smith chart. Now, we need to transform the zero reflection coefficient (normalized impedance of 1) to S (point D) at the input and to L (point B) at the output. One way to achieve this is to move first on the unity conductance circle from the center of the chart to point C and then on the constant-resistance circle to reach point D. Thus, it requires a shunt capacitor and then a series inductor to match at the input. Similarly, we can follow the unity-resistance circle from 1 to point A and then the constant-conductance circle to reach point B. Thus, a capacitor is connected in series with the load and then an inductor in shunt to match its output side. Actual values of the components are determined as follows. The normalized susceptance at point C is estimated as j 1.7. Hence, the shunt capacitor on the source side must provide a susceptance of 1.7/50 = 0.034 S.

431

AMPLIFIER DESIGN FOR MAXIMUM GAIN

0

110

90

100

80

70 60

12

50

30

20

160

15

30

0

14

40

0

1

170

10

D B

10

5

2

1

0.5

−10

70 0 −1 −16

0.2

0

180

0

A 0

−1

−4

40

−

0

−3

0 15

−20

C

−5

30

0

−6

0

−70

20

−80

0

−90

−100 −11

−1

−1

Figure 11.4

Smith chart illustrating the design of Example 11.2.

2.307 pF

50 Ω

5.1725 nH 12.434 nH

~

50 Ω

5.411 pF

Figure 11.5 RF circuit designed for Example 11.2.

Since the signal frequency is 1 GHz, this capacitance must be 5.411 pF. Now the change in normalized reactance from point C to point D is determined as j 0.2 − (−j 0.45) = j 0.65. The positive sign indicates that it is an inductor of reactance 0.65 × 50 = 32.5 . The corresponding inductance is found as 5.1725 nH. Similarly, the normalized reactance at point A is estimated as −j 1.38. Hence, the load requires a series capacitor of 2.307 pF. For transforming the susceptance of point A to that of point B, we need an inductance of (−0.16 − 0.48)/50 = −0.0128 S. It is found to be a shunt inductor of 12.434 nH. This circuit is illustrated in Figure 11.5.

432

TRANSISTOR AMPLIFIER DESIGN

Bilateral Case (Simultaneous Conjugate Matching) To obtain the maximum possible gain, a bilateral transistor must be matched at both of its ports simultaneously, as shown in Figure 11.6. When its output port is properly terminated, the input side of the transistor is matched with the source such that in = S∗ . Similarly, the output port of transistor is matched with the load when its input is matched terminated. Mathematically, in = S∗ and

out = L∗

Therefore, S∗ = S11 +

S12 S21 L 1 − S22 L

(11.2.1)

L∗ = S22 +

S12 S21 S 1 − S11 S

(11.2.2)

∗ ∗ S12 S21 ∗ ∗ (1/ L ) − S22

(11.2.3)

and

From (11.2.1), ∗ + S = S11

and from (11.2.2), L∗ =

S22 − (S11 S22 − S12 S21 )S S22 − S = 1 − S11 S 1 − S11 S

(11.2.4)

Substituting (11.2.4) in (11.2.3), we have ∗ + S = S11

∗ ∗ S12 S21 ∗ [(1 − S11 S )/(S22 − S )] − S22

ZS

VS ∠0°

~

Input matching network

Output match- Load ing ZL network

Transistor

ΓS

Γin

Γout

ΓL

Figure 11.6 Bilateral transistor with input- and output-matching networks.

433

AMPLIFIER DESIGN FOR MAXIMUM GAIN

or ∗ + S = S11

∗ ∗ ∗ ∗ S12 S12 S21 (S22 − S ) S21 (S22 − S ) ∗ + = S 11 ∗ ∗ 1 − S11 S − |S22 |2 + S S22 1 − |S22 |2 − (S11 − S22 )S

or ∗ ∗ ∗ ∗ (1 − |S22 |2 )S − (S11 − S22 )S2 = S11 (1 − |S22 |2 ) − S11 (S11 − S22 )S ∗ ∗ + S12 S21 (S22 − S )

or ∗ ∗ (S11 − S22 )S2 + (||2 − |S11 |2 + |S22 |2 − 1)S + (S11 − S22 ∗ ) = 0

This is a quadratic equation in S and its solution can be found as follows: S =

B1 ±

B12 − 4|C1 |2 2C1

= MS

(11.2.5)

where B1 = 1 + |S11 |2 − |S22 |2 − ||2 and

∗ C1 = S11 − S22

Similarly, a quadratic equation for L can be formulated. Solutions to that equation are found to be ML =

B2 ±

B22 − 4|C2 |2 2C2

(11.2.6)

where B2 = 1 + |S22 |2 − |S11 |2 − ||2 and ∗ C2 = S22 − S11

If |B1 /2C1 | > 1 and B1 > 0 in equation (11.2.5), the solution with a minus sign produces |MS | < 1 and that with a positive sign produces |MS | > 1. On other hand, if |B1 /2C1 | > 1 with B1 < 0 in this equation, the solution with a plus sign produces |MS | < 1 and the solution with a minus sign produces |MS | > 1. Similar considerations apply to equation (11.2.6) as well. Observation

|Bi /2Ci | > 1 implies that |k| > 1.

434

TRANSISTOR AMPLIFIER DESIGN

Proof: Bi 2C

i

or

2 2 2 > 1 ⇒ 1 + |S11 | − |S22 | − || > 1 2(S − S ∗ ) 11

22

∗ |1 + |S11 |2 − |S22 |2 − ||2 | > 2|(S11 − S22 )|

Squaring on both sides of this inequality, we get ∗ |1 + |S11 |2 − |S22 |2 − ||2 |2 > 4|(S11 − S22 )|2

and ∗ ∗ ∗ |S11 − S22 |2 = (S11 − S22 )(S11 − S22 ∗ )

= |S12 S21 |2 + (1 − |S22 |2 )(|S11 |2 − ||2 ) Therefore, |1 + |S11 |2 − |S22 |2 − ||2 |2 > 4|S12 S21 |2 + 4(1 − |S22 |2 )(|S11 |2 − ||2 ) or |1 + |S11 |2 − |S22 |2 − ||2 |2 − 4(1 − |S22 |2 )(|S11 |2 − ||2 ) > 4|S12 S21 |2 or (1 − |S22 |2 − |S11 |2 + ||2 )2 > 4|S12 S21 |2 or |1 − |S22 |2 − |S11 |2 + ||2 | > 2|S12 S21 | Therefore,

|1 − |S22 |2 − |S11 |2 + ||2 | >1⇒k>1 2|S12 S21 |

Also, it can be proved that || < 1 implies that B1 > 0 and B2 > 0. Therefore, minus signs must be used in equations (11.2.5) and (11.2.6). Since GT max =

(1 − |MS |2 )|S21 |2 (1 − |ML |2 ) |(1 − S11 MS )(1 − S22 ML ) − S12 S21 ML MS |2

substituting equations (11.2.5), (11.2.6), and k from (11.1.18) gives GT max =

|S21 | (k − k 2 − 1) |S12 |

(11.2.7)

435

AMPLIFIER DESIGN FOR MAXIMUM GAIN

Maximum stable gain is defined as the value of GT max when k = 1. Therefore, S21 GMSG = S12

(11.2.8)

Example 11.3 The S-parameters of a properly biased GaAs FET HFET-1101 are measured using a 50- network analyzer at 6 GHz as follows: S11 = 0.614 −167.4◦ , S21 = 2.187 32.4◦ , S12 = 0.046 65◦ , and S22 = 0.716 −83◦ . Design an amplifier using this transistor for a maximum possible gain. SOLUTION First we need to test for stability using (11.1.16) and (11.1.18). || = |S11 S22 − S12 S21 | = 0.3419 and k=

1 − |S11 |2 − |S22 |2 + ||2 = 1.1296 2|S12 S21 |

Since both of the conditions are satisfied, the transistor is unconditionally stable. For the maximum possible gain, we need to use simultaneous conjugate matching. Therefore, the source and load reflection coefficients are determined from (11.2.5) and (11.2.6) as follows:

MS =

B1 −

2C1

and ML =

B12 − 4|C1 |2

B2 −

B22 − 4|C2 |2 2C2

◦

= 0.8673 169.76

◦

= 0.9011 84.48

There are many ways to synthesize these circuits. The design of one of the circuits is discussed here. As illustrated in Figure 11.7, normalized impedance points A and C are identified corresponding to MS and ML , respectively. The respective normalized admittance points B and D are located next. One way to transform the normalized admittance of unity to that of point B is to add a shunt susceptance (normalized) of about j 2.7 at point E. This can be achieved with a capacitor or a stub of 0.194λ, as shown in Figure 11.8. Now, for moving from point E to point B, we require a transmission line of about 0.065λ. Similarly, an open-circuited shunt stub of 0.297λ will transform the unity value to 1 −j 3.4 at point F, and a transmission line length of 0.093λ will transform it to the desired value at point D. Note that both of the shunt stubs are asymmetrical about the main line. A symmetrical stub is preferable. It can be achieved via two shunt-connected stubs of susceptance j 1.35 at the input end and of −j 1.7 at the output. This circuit

436

TRANSISTOR AMPLIFIER DESIGN

110

0 12

90

100

80

70 60

C

50

30

15

30

0

40

14 0

1

160

10

170

20

E

A 5

2

1

0.5

10

−10

70 0 −1

0.2

0

180

0

B

−

0

−4

0 14

−1

−

50

−16

−20 30

F

−5

30

0

−6

0

−70

−80

−90

20

0

−100 − 11

−1

−1

Figure 11.7

D

Design of input- and output-side matching networks.

0.093λ

0.065λ 50 Ω

0.297λ 50 Ω O.C. 0.194λ

O.C.

Figure 11.8

RF part of the amplifier circuit for Example 11.3.

437

AMPLIFIER DESIGN FOR MAXIMUM GAIN O.C. 0.333λ 0.093λ

O.C. 0.15λ 0.065λ

50 Ω

0.333λ 50 Ω O.C.

0.15λ

O.C.

Figure 11.9

RF part of the amplifier circuit with symmetrical stubs.

is illustrated in Figure 11.9. The value of the gain is evaluated from (11.2.7) as follows: GT max =

|S21 | (k − k 2 − 1) = 28.728 ⇒ 10 log(28.728) = 14.58 dB |S12 |

Unilateral Figure of Merit In preceding examples we found that the design with a bilateral transistor is a bit complex compared with a unilateral case. The procedure for the bilateral transistor becomes even more cumbersome when the specified gain is less than its maximum possible value. It can be simplified by assuming that a bilateral transistor is unilateral. The unilateral figure of merit provides an estimate of the error associated with this assumption. It can be formulated from (10.4.8) and (10.4.10) as follows: 2 GT |1 − S22 L |2 1 − S22 L = = 2 GT U |1 − out L | 1 − S22 L − S12 S21 S L /(1 − S11 S ) 2 1 |1 − S22 L |2 = 1 − S S /(1 − S )(1 − S ) |1 − out L |2 12 21 S L 22 L 11 S or

1 2 GT = GT U 1 − X

where X=

S12 S21 S L (1 − S22 L )(1 − S11 S )

438

TRANSISTOR AMPLIFIER DESIGN

Therefore, the bounds of this gain ratio are given by 1 GT 1 < < 2 (1 + |X|) GT U (1 − |X|)2 ∗ ∗ When S = S11 and L = S22 , GT U has a maximum value. In this case, the maximum error introduced by using GT U in place of GT ranges as follows:

GT 1 1 < < (1 + U )2 GT U (1 − U )2 where U=

|S12 ||S21 ||S11 ||S22 | (1 − |S11 |2 )(1 − |S22 |2 )

(11.2.9)

(11.2.10)

The parameter U is known as the unilateral figure of merit. Example 11.4 The scattering parameters of two transistors are as given below. Compare the unilateral figures of merit of the two. For transistor A, S11 = 0.45 150◦ , S12 = 0.01 −10◦ , S21 = 2.05 10◦ , and S22 = 0.4 −150◦ . For transistor B, S11 = 0.641 −171.3◦ , S12 = 0.057 16.3◦ , S21 = 2.058 28.5◦ , and S22 = 0.572 −95.7◦ . SOLUTION From (11.2.10) we find that U=

0.01 × 2.05 × 0.45 × 0.4 = 0.00551 = UA (1 − 0.452 )(1 − 0.42 )

for transistor A. Similarly, for transistor B, U=

0.057 × 2.058 × 0.641 × 0.572 = 0.1085 = UB (1 − 0.6412 )(1 − 0.5722 )

Hence, the error bounds for these two transistors can be determined from (11.2.9) as follows. For transistor A, 0.9891

0 vL < 0

(13.3.8)

This s(t) can be expressed in terms of the Fourier series as ∞

s(t) =

1 2 nπ + sin cos nωL t 2 n=1 nπ 2

(13.3.9)

For vi = V cos ωi t, the output voltage is found to be ∞ V V 1 nπ cos ωi t + sin {cos[(nωL + ωi )t] + cos[(nωL − ωi )t]} 2 π n=1 n 2 (13.3.10) Thus, the output of this mixer circuit does not contain the local oscillator signal. However, it includes the input signal frequency ωi that appears with the sidebands and spurious signals. It can be a problem in certain applications. For example, consider a case where the input and the local oscillator signal frequencies at a receiver are 60 and 90 MHz, respectively. As illustrated in Figure 13.24, the difference signal frequency at its output is 30 MHz. If a 30-MHz signal is also

vo =

563

SWITCHING-TYPE MIXERS

60 MHz

30 MHz

Filter and IF section

90 MHz

Figure 13.24 Single-balanced mixer of Figure 13.22 at the receiver.

present at the input, it will get through the mixer as well. A preselector (bandpass filter) may be used to suppress the undesired input. Note from (13.3.10) that the amplitude of the sidebands is only V /π, whereas it was twice this in (13.3.6). It influences the sensitivity of the receiver employing this mixer. In the following section we consider another mixer circuit, which solves most of these problems. Double-Balanced Mixer A double-balanced mixer circuit requires four diodes and two transformers with secondary sides center-tapped, as shown in Figure 13.25. As before, it is assumed that the local oscillator voltage vL is high compared with the input vi . Hence, diodes D1 and D2 conduct when the local oscillator voltage is positive while the other two are in the off state. This situation turns the other way around when the voltage vL reverses its polarity. Thus, diodes D1 and D2 conduct for the positive half cycle of vL and D3 and D4 conduct for the negative half. Equivalent circuits can be drawn separately for these two states. Figure 13.26 shows one such equivalent circuit, which represents the case of positive vL . Diode resistance in the forward-biased condition is assumed to be rd . The equivalent circuit can be simplified further to facilitate the analysis. Consider the simplified equivalent circuit shown in Figure 13.27. The currents in the two loops are i1 and i2 . Using Kirchhoff’s voltage law, loop equations can

D4

R

RG + vG −

+ vi −

+ vi − + vi −

D2

+ vL

D3

− D1 RL

Figure 13.25

vo

Switching-type double-balanced mixer.

Local oscillator

564

DETECTORS AND MIXERS R

RG +

+ vi −

vG −

rd

+ vi − + vi −

+ Local oscillator

vL −

rd RL

vo

Figure 13.26 Equivalent of the double-balanced mixer with positive vL . rd + −

vi

+

vL

i1

−

RL

+

i2

vL − rd

Figure 13.27 Simplified equivalent circuit of Figure 13.26.

be written as vi = RL (i1 − i2 ) + rd i1 − vL

(13.3.11)

vi = RL (i1 − i2 ) − rd i2 + vL

(13.3.12)

and

Equations (13.3.11) and (13.3.12) can be solved for i1 − i2 . Hence, i1 − i2 =

vi RL + rd /2

But from Ohm’s law, i1 − i2 = −

vo RL

(13.3.13)

(13.3.14)

Combining (13.3.13) and (13.3.14) yields vo RL =− vi RL + rd /2

(13.3.15)

CONVERSION LOSS

565

Similarly, an equivalent circuit can be drawn for the negative half cycle of vL . Diodes D3 and D4 are conducting in this case, and following an identical analysis it may easily be found that RL vo =+ vi RL + rd /2

(13.3.16)

Equations (13.3.15) and (13.3.16) can be expressed by a single equation after using the function s(t) as defined in (13.3.3). Hence, vo =

RL RL vi = vi s(t) RL + rd /2 RL + rd /2

(13.3.17)

Now using the Fourier series representation for s(t) as given in (13.3.4), (13.3.17) can be expressed as ∞

RL 4 (−1)n+1 nπ vo = vi sin cos nωL t RL + rd /2 π n=1 n 2

(13.3.18)

In case of a sinusoidal input, vi = V cos ωi t, the output voltage vo can be found as vo =

∞ RL 2V (−1)n+1 nπ sin {cos[(nωL + ωi )t] RL + (rd /2) π n=1 n 2

+ cos[(nωL − ωi )t]}

(13.3.19)

Hence, this mixer circuit produces the upper and lower sidebands along with an infinite number of spurious signals. However, the ωL and ωi signals are both isolated from the output. This analysis assumes that the diodes are perfectly matched and the transformers are ideal. Variations of these characteristics may not perfectly isolate the output from the input and the local oscillator. These double-balanced mixers are commercially available for applications in the audio through microwave frequency bands.

13.4 CONVERSION LOSS Since mixers are used to convert the frequency of an input signal, a circuit designer would like to know if there is some change in its power as well. This information is especially important in receiver design where the signal received may be fairly weak. Conversion loss (or gain) of the mixer is defined as a ratio of the power output in one of the sidebands to the power of its input signal. It can be evaluated as follows. Consider the double-balanced mixer circuit shown in Figure 13.24. Using its equivalent circuit illustrated in Figure 13.26, the input resistance Rin can be found

566

DETECTORS AND MIXERS

with the help of (13.3.13) as follows: Rin =

vi rd ≈ RL = RL + i1 − i2 2

(13.4.1)

It is assumed here that load resistance RL is much larger than the forward resistance rd of the diode. If there is maximum power transfer from the voltage source vG at the input to the circuit, the source resistance RG must be equal to the input resistance Rin . Since the input resistance is equal to the load RL , the following condition must be satisfied: RG = RL If the sinusoidal source voltage vG has a peak value of VP , it divides between two equal resistances RG and Rin , and only half of voltage VP appears across the transformer input. Hence, V2 Pi = P (13.4.2) 8RL From (13.3.19), the peak output voltage Vo of the sideband is Vo |ωL ±ωi =

2V VP = π π

(13.4.3)

Therefore, the output power Po is found as Po =

VP2 2π2 RL

(13.4.4)

If G is defined as the conversion gain, which is a ratio of its output to input power, then from (13.4.2) and (13.4.4) we find that G=

Po VP2 8RL 4 = = 2 Pi 2π2 RL VP2 π

(13.4.5)

Since G is less than unity, it is really not a gain but a loss of power. Hence, the conversion loss, L, is L = 10 log

π2 = 3.92 dB ≈ 4 dB 4

(13.4.6)

Hence, approximately 40% of the power of the input signal is transferred to the output of an ideal double-balanced mixer. Circuit loss and mismatch reduce it further. In the case of a single-balanced mixer where the local oscillator is isolated from the output, the sideband voltage can be found from (13.3.10). Hence, the

INTERMODULATION DISTORTION IN DIODE-RING MIXERS

power output in one of the sidebands is given as 2 1 VP2 V Po = = 2RL π 8π2 RL

567

(13.4.7)

Since the input power is still given by (13.4.2), its conversion gain may be found as Vs2 8RL 1 Po = × 2 = 2 (13.4.8) G= 2 Pi 8π RL Vs π Again, the output is less than the input power, and therefore, it represents a loss of signal power. The conversion loss for this case is L = 10 log π2 = 9.943 ≈ 10 dB

(13.4.9)

Hence, the conversion loss in this single-balanced mixer is 6 dB (four times) larger than that of the double-balanced mixer considered earlier. It may be noted that the mixer circuit shown in Figure 13.19 has a conversion loss of only 4 dB. These results show that the maximum possible power transferred from the input to the output of a diode mixer is less than 40%. Therefore, these circuits always have conversion loss. A conversion gain is possible only when a transistor (BJT or FET) circuit is used as the mixer. FET mixers are described briefly later in the chapter.

13.5 INTERMODULATION DISTORTION IN DIODE-RING MIXERS As discussed in Chapter 2, the noise level defines the lower end of the dynamic range of a system, whereas the distortion determines its upper end. Therefore, a circuit designer needs to know the distortion introduced by the mixer and whether there is some way to limit that. We consider here the mixer circuit depicted in Figure 13.28, which has a resistance R in series with each diode. D4

R

RG +

+ vi −

vG −

+ vi −

D2

R

+ vL

D3 R

+ vi −

R

v

−

D1 RL

vo

Figure 13.28 Mixer circuit with resistance R in each branch.

Local oscillator

568

DETECTORS AND MIXERS D1

R

Ro

iL − i + vL −

+ vo − iL + i

+ vi −

D2

RL

v

R

Figure 13.29 Simplified equivalent of the circuit shown in Figure 13.28 for vL > 0.

Assuming that vL is large enough that the local oscillator alone controls the conduction through the diodes, an equivalent circuit can be drawn for each half cycle. In the case of a positive half period of vL , the simplified equivalent circuit may be drawn as illustrated in Figure 13.29. Diodes D1 and D2 conduct for this duration while the other two are in the off state. Assume that iL is current due to vL and i is due to vi . Hence, currents through the diodes are given as follows: current through diode D1 = id3 = iL − i current through diode D2 = id4 = iL + i Since the V –I characteristic of a diode is given as id = Is (eαvd − 1) ≈ Is eαvd where α= we find that vd =

(13.5.1)

q nkT

(13.5.2)

1 id ln α Is

(13.5.3)

Note that net current flowing through the load resistance RL is the difference of two loop currents: iL − i and iL + i. Hence, the LO current through RL cancels out, and two components of current due to vi are added to make it 2i. Loop equations for the circuit can be found to be vL = (iL − i)R + vd3 + vi − RL · 2i

(13.5.4)

vL = RL · 2i − vi + R(iL + i) + vd4

(13.5.5)

and

Subtracting (13.5.5) from (13.5.4), we have 0 = −2(R + 2RL )i + 2vi + vd3 − vd4

569

INTERMODULATION DISTORTION IN DIODE-RING MIXERS

or

1 iL + i 1 iL + i iL − i 2(R + 2RL )i − 2vi = − ln = − ln ln α Is Is α iL − i

or vi = (R + 2RL )i +

1 iL + i ln 2α iL − i

(13.5.6)

Since the LO current iL is very large compared with i, it can be expanded as an infinite series: 1 1 5 1 3 i + i + ··· (13.5.7) vi = R + 2RL + i+ 3 αiL 3αiL 5αiL5 An inverse series solution to this can easily be found with the help of computer software (such as Mathematica) as follows: i=

vi3 vi − + ··· R + 2RL + 1/αiL 3αiL3 (R + 2RL + 1/αiL )4

(13.5.8)

Therefore, the output voltage vo can be written as vi3 vi vo = 2RL i = 2RL − + ··· R + 2RL + 1/αiL 3αiL3 (R + 2RL + 1/αiL )4 (13.5.9) On comparing it with (2.5.31), we find that the term responsible for the intermodulation distortion (IMD) is

k3 = −

3αiL3

2RL (R + 2RL + 1/αiL )4

(13.5.10)

Since the IMD is directly proportional to the cube of k3 , (13.5.10) must be minimized to reduce the intermodulation distortion. One way to achieve this objective is to use a large local oscillator current iL . Another possible way is to employ the resistance R in series with each diode. Sometimes an additional diode is used in place of R. This permits higher local oscillator drive levels and a corresponding reduction in IMD. Example 13.4 In the mixer circuit shown in Figure 13.30, transformer and diodes are ideal. The transformer has 400 turns on its primary and 800 turns with a center tap on its secondary side. Find the output voltage and the value of load RL that provides maximum power transfer from vs . SOLUTION For maximum power transfer, the impedance when transformed to the primary side of the transformer must be equal to 50 . Since the number of turns on either side of the center tap is the same as on its primary side and only one of the diodes is conducting at a given time, the load resistance RL must be

570

DETECTORS AND MIXERS

50 Ω

RL

vi vs

vo

vL

vi

RL

Figure 13.30 Mixer circuit for Example 13.4.

50 as well. Under this condition, only half of the signal voltage appears across the primary. Hence, vi = vs /2. Assume that vL is large compared with vi such that diode switching is controlled solely by the local oscillator. Therefore, v + vi vo = L −(vL + vi )

for vL > 0 for vL < 0

This output voltage can be expressed as follows: vo = (vL + vi )s(t) where

1 vL > 0 s(t) = −1 vL < 0

We already know the Fourier series representation for this s(t), given by (13.3.4) as ∞ nπ 4 (−1)n+1 sin cos nωL t s(t) = (13.3.4) π n=1 n 2 If vi and vs are sinusoidal voltages vi = Vi cos ωi t and vL = VL cos ωL t then the output voltage will be ∞

vo = (Vi cos ωi t + VL cos ωL t)

nπ 4 (−1)n+1 sin cos nωL t π n=1 n 2

571

FET MIXERS

or ∞

vo =

2 (−1)n+1 nπ sin {Vi [cos(nωL + ωi )t + cos(nωL − ωi )t] π n=1 n 2 + VL [cos(nωL + ωL )t + cos(nωL − ωL )t]}

13.6 FET MIXERS Diodes are commonly used in mixer circuits because of certain advantages, including a relatively low noise. A major disadvantage of these circuits is the conversion loss, which cannot be reduced below 4 dB. On the other hand, transistor circuits can provide a conversion gain, although they may be noisier. A well-designed FET mixer produces less distortion than do BJT circuits. Further, the range of its input voltage is generally much higher (10 times or so) with respect to BJT mixers. The switching and resistive types of mixer have been designed using the FET. In this section we present an overview of selected FET mixer circuits. Figure 13.31 shows a FET mixer circuit. Local oscillator signal vL is applied to the source terminal of the FET via capacitor C1 while the input v2 is connected to its gate. For the FET operating in saturation, we can write Vgs 2 (13.6.1) iD = IDSS 1 − Vp where iD is the drain current of the FET, Vgs the gate-to-source voltage, Vp its pinch-off voltage, and IDSS the drain current for Vgs = 0. For the circuit shown in Figure 13.31, the total gate voltage Vgs is Vgs = vi − vL + VGS VDD vo

C2

C1 vi

RG

RS vL

Figure 13.31

FET mixer circuit.

572

DETECTORS AND MIXERS

where VGS is the gate-to-source dc bias voltage. Therefore, the drain current iD in this case is vi − vL + VGS 2 iD = IDSS 1 − (13.6.2) Vp For sinusoidal vi and vL given as vi = Vi cos ωi t and vL = VL cos ωL t the drain current iD can be found from (13.6.2). Hence, Vi cos ωi t − VL cos ωL t + VGS 2 iD = IDSS 1 − Vp or 2 [Vi cos ωi t − VL cos ωL t + VGS ] iD = IDSS 1 − Vp +

IDSS 2 [V + 2VGS Vi cos ωi t + Vi2 cos2 ωi t − 2VL VGS cos ωL t Vp2 GS + VL2 cos2 ωL t − 2Vi VL cos ωi t cos ωL t]

(13.6.3)

Equation (13.6.3) can be rearranged in the following form and then the amplitudes of sum and difference frequency components can easily be identified as follows: iD = IDC + a1 cos ωi t + a2 cos ωL t + b1 cos 2 ωi t + 2 cos 2 ωL t − c{cos[(ωL + ωi )t] + cos[(ωL − ωi )t]} where c=

IDSS Vi VL Vp2

(13.6.4)

The ratio of amplitude c to input voltage Vi is defined as the conversion transconductance gc . Hence, c IDSS VL = (13.6.5) gc = Vi Vp2 For high conversion gain, gc must be large. It appears from (13.6.5) that the FET with high IDSS should be preferable, but that is generally not the case. IDSS and Vp of a FET are related such that the device with high IDSS also has high Vp . Therefore, it may result in a lower gc than that of a low IDSS device. VL

573

FET MIXERS

is directly related to the conversion gain. Hence, higher local oscillator voltage increases the conversion gain. Since the FET is to be operated in its saturation mode (the constant-current region), VL must be less than the magnitude of the pinch-off voltage. As a special case, if VL = |Vp |/2, the conversion transconductance may be found as IDSS gc = (13.6.6) 2Vp Since the transconductance gm of a JFET is given as gm =

∂iD 2IDSS =− ∂VGS Vp

1−

VGS Vp

(13.6.7)

and gm |VGS =0 = −

2IDSS Vp

(13.6.8)

the conversion transconductance is one-fourth of the small-signal transconductance evaluated at VGS = 0 (provided that VL = 0.5Vp ). For a MOSFET, it can be shown that the conversion transconductance cannot exceed one-half of the small-signal transconductance. Note from (13.6.3) that there are fundamental terms of input as well as local oscillator frequencies, their second-order harmonics, and the desired ωi ± ωL components. Unlike the diode mixer, higher-order spurious signals are not present in this output. In reality, there is some possibility for those terms to be present because of the circuit imperfections. An alternative to the circuit shown in Figure 13.31 combines the local oscillator signal with the input before applying it to the gate of FET. The analysis of this circuit is similar to the one presented above. In either case, one of the serious problems is isolation among the three signals. There have been numerous attempts to address those problems. Example 13.5 A given JFET has IDSS = 50 mA and transconductance gm = 200 mS when its gate voltage VGS is zero. If it is being used in a mixer that has a 50- load, find the conversion gain of the circuit. SOLUTION Since gm |VGS =0 = − then Vp =

2IDSS Vp

2 × 50 × 10−3 1 = = 0.5 V 200 × 10−3 2

574

DETECTORS AND MIXERS

If the local oscillator voltage VL is kept at approximately 0.25 V, the conversion transconductance gc may be found as gc =

50 × 10−3 = 50 × 10−3 S = 50 mS 2 × 0.5

Therefore, the magnitude of voltage gain Av is Av ≈ gc RL = 50 × 10−3 × 50 = 2.5 Since a common-gate configuration of the FET is generally used in the mixer circuit, its current gain is close to unity. Therefore, the conversion power gain of this circuit is approximately 2.5. A dual-gate FET is frequently employed in a mixer circuit. There are several circuit configurations and modes of operation for dual-gate FET mixers. Figure 13.32 illustrates a circuit that usually performs well in most receiver designs. The working of this circuit can be explained considering that a dualgate FET represents two single-gate transistors connected in cascode. Hence, the upper FET works as a source follower and the lower one provides mixing. An RF signal is applied to a gate that is close to the ground while the local oscillator is connected at the other gate (one close to the drain in common-source configuration). If RF is connected to the other gate, the drain resistance of the lower transistor section appears as source resistance and thus reduces the signal. This circuit is used as a mixer when both transistors are operating in their triode (nonsaturation) mode. A bypass circuit is needed before its output to block the input (RF) and the local oscillator signals. Further, another bypass circuit may be needed across the local oscillator to block the output frequencies. This circuit has lower gain and higher noise, but a properly designed mixer can achieve a modestly higher third-harmonic intercept compared with a single-gate mixer. Figure 13.33 shows a FET mixer circuit that employs two transistors and transformers. A local oscillator signal is applied to both transistors in the same phase while one RF input is out of phase with respect to the other. The resulting output is a differential of the two sides, and therefore the local oscillator signal is

Local oscillator vL

Matching network

RF input vi

Matching network

Matching network

Figure 13.32 Dual-gate FET mixer.

Output vo

575

FET MIXERS

vi

vo

+ VDD

vL

RL

Rs

Figure 13.33 FET balanced mixer.

canceled out while the RF is added. The nonlinear characteristic of the transistors generates the upper and lower sideband signals. If the dc bias voltage at the source is VS , the RF input is Vi cos ωi t, and the local oscillator voltage is VL cos ωL t, the output voltage of this circuit is vo =

4RL Vi 4RL Vi VS + VP VP2

cos ωi t +

2RL Vi VL VP2

× [cos(ωL + ωi )t + cos(ωL − ωi )t]

(13.6.9)

This shows that the output of this circuit contains the RF input frequency along with two sidebands. Asymmetry and other imperfection can add more frequency components as well. Figure 13.34 illustrates an active double-balanced circuit, known as the Gilbert cell mixer, commonly employed in radio-frequency integrated circuits (RFICs). Transistors Q1 and Q2 are used to convert the RF voltage to current, and Q3 to Q6 are used for the switching. RF and local oscillator signals are applied to the circuit via the corresponding matching networks (baluns). Similarly, the IF output requires a matching network as well. A detailed analysis of the circuit is beyond the scope of this book; only a qualitative analysis is presented here. Assume that the local oscillator signal is a large square wave that can turn on the transistors when it is high and turn them off when it is low. Note that the current iif1 is the sum of i3 and i5 , and iif2 is the sum of i4 and i6 . Further, the RF current i1 is the sum of i3 and i4 and the sum of i5 and i6 is equal to i2 . The bias current Ib is the sum of i1 and i2 . Since a transistor is on only when the local oscillator signal is high, transistors Q4 and Q5 will be on while Q3 and Q6 will be in off conditions. Therefore, iif1 = i5 = i2 and iif2 = i4 = i1

576

DETECTORS AND MIXERS

VDD RD RD iif2

iif1

Balun

LO

i3 Q3

Balun i5

i4 Q4

Q6

i2

Q1 RF

i6

Q5

i1

IF

Q2

Balun Ib

Figure 13.34 Gilbert cell mixer circuit.

Similarly, when the local oscillator signal is low, transistors Q3 and Q6 will be on and Q4 and Q5 will be off. Therefore, iif1 = i3 = i1 and iif2 = i6 = i2 Hence the IF output vif can be found to be vif =

RD (i2 − i1 ) RD (i1 − i2 )

vL > 0 vL < 0

(13.6.10)

or vif = RD (i2 − i1 )s(t)

(13.6.11)

where s(t) is defined by (13.3.3) and (13.3.4). Further, the conversion gain Gc of this circuit can be found after assuming that it is symmetrical, and therefore the transconductance gm of Q1 is the same as that of Q2 . It is given as Gc =

4 gm RD π

(13.6.12)

577

PROBLEMS

SUGGESTED READING Bahl, I. J., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Larson, L. E. (ed.), RF and Microwave Circuit Design for Wireless Communications. Boston: Artech House, 1996. Leung, B., VLSI for Wireless Communication. Upper Saddle River, NJ: Prentice Hall, 2002. Maas, S. A., Microwave Mixers. Boston: Artech House, 1993. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Rohde, U. L., Microwave and Wireless Synthesizers. New York: Wiley, 1997. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1998. Vendelin, G. D., A. Pavio, and U. L. Rhode, Microwave Circuit Design Using Linear and Non-linear Techniques. New York: Wiley, 1990.

PROBLEMS 13.1. In an AM signal, the carrier output is 1 kW. If the wave modulation is 100%, determine the power in each sideband. How much power is being transmitted? 13.2. Assume that an AM transmitter is modulated with a video signal given by m(t) = −0.15 + 0.7 sin ω1 t, where f1 is 4 MHz. Let the unmodulated carrier amplitude be 100 V. Evaluate and sketch the spectrum of modulated signal. 13.3. An AM broadcast station that uses 95% modulation is operating at total output power of 100 kW. Find the power transmitted in its sidebands. 13.4. With a modulation index of 80% an AM transmitter produces 15 kW. How much of this is carrier power? Find the percentage power saving if the carrier and one of the sidebands are suppressed before transmission. 13.5. When an AM station transmits an unmodulated carrier, the current through its antenna is found to be 10 A. It increases to 12 A with a modulated signal. Determine the modulation index used by the station. 13.6. In the circuit arrangement shown in Figure P13.6, Vout = V1 − V2 is the output signal, where V1 = 10VC + 0.2VC Vm and V2 = 6VC − 0.2VC Vm .

Vout VC = C cos ωct

V1

V2 Vm = M cos ωmt

Figure P13.6

578

DETECTORS AND MIXERS

(a) Calculate the value of M that results in a modulation index of 0.8. (b) Determine the value of C that results in an unmodulated output voltage amplitude of 8 V. (c) If this Vout is applied to a 50- load, find the power in the carrier and in each sideband. 13.7. The circuit shown in Figure P13.7 uses a zero-biased diode with Rj = 500 , and n = 1 at 290 K. If the signal applied to this circuit is vin (t) = 0.5[1 + 0.3 cos(4π × 104 t + 0.15)] cos(π × 109 t − 0.15)

V

find the output voltage vout . Also, sketch the frequency spectra of vin and vout .

1 kΩ

31.83 pF vout

vin

Figure P13.7

13.8. A 1-MHz carrier signal is amplitude-modulated simultaneously with 300Hz, 800-Hz, and 2-kHz sinusoidal signals. What frequencies are present in the output? 13.9. An AM broadcast transmitter radiates 50 kW of carrier power. If the modulation index is 0.85, find the power radiated. 13.10. An FM broadcast station uses a 10-kHz audio signal of 2 V peak value to modulate the carrier. If the allowed frequency deviation is 20 kHz, determine the bandwidth requirement. 13.11. A frequency modulator is supplied with a carrier of 10 MHz and a 1.5kHz audio signal of 1 V. The frequency deviation at its output is found to be ±9 kHz. (a) An FM wave from the modulator is passed through a series of frequency multipliers with a total multiplication of 15 (i.e., fo = 15fc ). Determine the frequency deviation at the output of the multiplier. (b) An output of the modulator is mixed with a 58-MHz signal. Find the sum frequency output of the mixer and its frequency deviation. (c) Find the modulation index at the output of the modulator. (d) The audio input to the modulator is changed to 500 Hz with its amplitude at 2 V. Determine the modulation index and frequency deviation at the output of the modulator. 13.12. An FM signal, vFM (t) = 1500 cos[2π × 109 t + 2 sin(3π × 105 )]

V

579

PROBLEMS

is applied to a 75- antenna. Determine (a) the carrier frequency, (b) the transmitted power, (c) the modulation index, and (d) the frequency of the modulating signal. 13.13. When the modulating frequency in FM is 400 Hz and the modulating voltage is 2.4 V, the modulation index is 60. Calculate the maximum deviation in frequency. What is the modulation index when the modulating frequency is reduced to 250 Hz and the modulating voltage is simultaneously raised to 3.2 V? 13.14. The iD –Vgs characteristics of a FET are given by

2Vgs iD = 7 1 − 7

2 mA

If it is being used in a mixer that has a 50- load, find the conversion gain of the circuit. 13.15. Calculate the conversion loss of the double-balanced mixer shown in Figure P13.15. The diode ON resistance is much less than the load resistance RL . What value of RL is required for maximum power transfer if the transformer employed in the circuit has 400 turns in its primary and 800 turns with a center tap on its secondary side? 50 Ω

vi vs

vi

RL vo

vL RL

Figure P13.15

13.16. In the mixer circuit shown in Figure P13.16, the transformer and diodes are ideal. The transformer has 200 turns on each side. Find the value of RL that provides maximum power transfer from the source. 75 Ω

vi vs

vi

RL vo

vL RL

Figure P13.16

APPENDIX 1 DECIBELS AND NEPER

Consider the two-port network shown in Figure A1.1. Assume that V1 and V2 are the voltages at its ports. The voltage gain (or loss) Gv of this circuit is expressed in decibels as V2 Gv = 20 log10 dB (A1.1) V1 Similarly, if P1 and P2 are the power levels at the two ports, the power gain (or loss) of this circuit in decibels is given as Gp = 10 log10

P2 P1

dB

(A1.2)

Thus, the dB unit provides a relative level of the signal. For example, if we are asked to find P2 in watts for Gp as 3 dB, we also need P1 . Otherwise, the only information we can deduce is that P2 is twice P1 . Sometimes power is expressed in logarithmic units, such as dBW and dBm. These units are defined as follows. If P is power in watts, it can be expressed in dBW as G = 10 log10 P dBW (A1.3) On the other hand, if P is in milliwatts, the corresponding dBm power is found as G = 10 log10 P

dBm

(A1.4)

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

580

581

DECIBELS AND NEPER

Two-port network

V1

V2

Figure A1.1 Two-port network.

Thus, the dBW and dBm units represent power relative to 1 W and 1 mW, respectively. Another decibel unit that is commonly used to specify phase noise of an oscillator or strengths of various sidebands of a modulated signal is dBc. It specifies the signal strength relative to the carrier. Consider a 100-MHz oscillator that has an output power of −10 dBm. Suppose that its output power is −30 dBm in the frequency range 105 to 106 MHz; then power per hertz in the output spectrum is −90 dBm and the phase noise is −80 dBc. In general, if amplitudes of the sideband and the carrier are given as V2 and Vc , respectively, the sideband in dBc is found as G2 = 20 log

V2 Vc

(A1.5)

Consider now a 1-m-long transmission line with its attenuation constant α. If V1 is the signal voltage at its input port, the voltage V2 at its output is given as |V2 | = |V1 |e−α

(A1.6)

Therefore, the voltage gain Gv of this circuit in nepers is Gv (NP ) = ln

|V2 | = −α |V1 |

NP

(A1.7)

On the other hand, Gv in decibels is found to be Gv (dB) = 20 log

|V2 | = 20 log e−α = −20α log e = −8.6859α |V1 |

dB

Therefore, 1NP = 8.6859 dB

(A1.8)

The negative sign indicates that V2 is smaller than V1 and there is loss of signal.

APPENDIX 2 CHARACTERISTICS OF SELECTED TRANSMISSION LINES

COAXIAL LINE Consider a coaxial line with its inner and outer conductor radii a and b, respectively, as illustrated in Figure A2.1. Further, εr is the dielectric constant of the insulating material. The line parameters L, R, C, and G of this coaxial line are found as follows: 55.63εr C= pF/m (A2.1) ln(b/a) and L = 200 ln

b a

nH/m

(A2.2)

If the coaxial line has small losses due to imperfect conductor and insulator, its resistance and conductance parameters can be calculated as follows:

1 1 R ≈ 10 + a b

fGHz σ

/m

(A2.3)

and G=

0.3495εr fGHz tan δ ln(b/a)

S/m

(A2.4)

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

582

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

583

a

εr

b

Figure A2.1 Coaxial line geometry.

where tan δ is the loss tangent of the dielectric material, σ the conductivity of the conductors in S/m, and fGHz the signal frequency in GHz. The characteristic impedance and propagation constant of the coaxial line can be calculated easily using the formulas given in Chapter 3. Attenuation constants αc and αd due to conductor and dielectric losses, respectively, may be determined from Rs 1 1 αc = √ + (A2.5) b 2 µ0 /ε0 εr ln(b/a) a ω√ µ0 ε0 εr tan δ (A2.6) αd = 2 where ωµ0 Rs = (A2.7) 2σ STRIP LINE Strip line geometry is illustrated in Figure A2.2. Insulating material of thickness h has a dielectric constant εr . The width and thickness of the central conducting strip are w and t, respectively. For the case of t = 0, its characteristic impedance can be found as follows: √ −1 1 + x 296.1 0.6931 + ln 0 < x ≤ 0.7 √ 1 1 − x Z0 = √ (A2.8) √ εr x 1 + 30 0.6931 + ln √ 0.7 ≤ x < 1 1− x πw (A2.9) x = tanh 2h

584

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

w t

h

εr

Figure A2.2 Strip line geometry.

and x=

1 − x 2

(A2.10)

For the design of a strip line, the following convenient formulas can be used: √ w = 0.6366 tanh−1 k h

(A2.11)

where

[(eπ/y − 2)/(eπ/y + 2)]2

1 − [(eπy − 2)/(eπy + 2)]4 √ εr y = Z0 94.18 k=

0≤y≤1 1≤y

(A2.12) (A2.13)

For t = 0, w = 1 − 2 h

(A2.14) √

1 = 2.5465(1 − ) and

ey + 0.568 ey − 1

2 δ 0.0796 1 − 0.5 ln + 2 = π 2− 1 − 0.26

(A2.15)

(A2.16)

where = δ= and y is as defined in (A2.13).

t h

(A2.17) 2

1+

2 [/(1 3

− )]

(A2.18)

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

585

MICROSTRIP LINE The geometry of a microstrip line is illustrated in Figure A2.3. Dielectric substrate on the conducting ground is h meters high and its dielectric constant is εr . The width and thickness of the conducting strip on its top are w and t, respectively. The characteristic impedance of this line can be found as follows: 8h we we 60 for ≤1 √εre ln we + 4h h Z0 = w −1 376.7 we we e + 1.393 + 0.667 ln + 1.444 ≥1 for √ εre h h h (A2.19) where w w w t 1 + 0.3979 1 + ln 12.5664 ≤ h h t h 2π we (A2.20) = w t h w 1 h + 0.3979 1 + ln 2 ≥ h h t h 2π The effective dielectric constant εre of a microstrip line ranges between εr and 1 because of its propagation characteristics. If the signal frequency is low such that the dispersion is not a problem, it can be determined as follows: w ε − 1 t h r − εre = 0.5 εr + 1 + (εr − 1)F (A2.21) h 4.6 h w where w 2 h −0.5 1 + 12 + 0.04 1 − w w h = F −0.5 h h 1 + 12 w

w ≤1 h w ≥1 h

w t

h

εr

Figure A2.3

Microstrip line geometry.

(A2.22)

586

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

If dispersion cannot be ignored, the effective dielectric constant may be found as follows: √ 2 √ εr − εre √ εre (f ) = + εre (A2.23) 1 + 4F −1.5 where

40 w 2 F = fGHz h εr − 1 0.5 + 1 + 2 log 1 + 3 h

(A2.24)

The corresponding characteristic impedance is determined from the following formula: εre (f ) − 1 εre Z0 (f ) = Z0 (A2.25) εre − 1 εre (f ) Attenuation constants αc and αd for the conductor and dielectric losses, tively, are determined as follows: fGHz 32 − (we / h)2 ζ Np/m 9.9825 hZ0 σ 32 + (we / h)2 αc = 0.667(we / h) ζZ0 εre fGHz we −5 44.1255 × 10 + Np/m h σ h (we / h) + 1.444

respec-

w ≤1 h w ≥1 h (A2.26)

where

w 1.25 1.25t h 1 + we 1 + πw + π ln 4π t ζ= h 1.25t 1.25 h 1− + ln 2 1+ we πh π t

w 1 ≤ h 2π w 1 ≥ h 2π

(A2.27)

σ is conductivity of the conductor and fGHz is the signal frequency in GHz. αd = 10.4766

εr εre − 1 fGHz tan δ √ εr − 1 εre

Np/m

(A2.28)

For the design of a microstrip line that has negligible dispersion, the following formulas may be more convenient. For A > 1.52, w 8eA = 2A h e −2

(A2.29)

For A ≤ 1.52,

w 0.61 εr − 1 ln(B − 1) + 0.39 − = 0.6366 B − 1 − ln(2B − 1) + h 2εr εr (A2.30)

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

587

where A=

εr − 1 Z0

0.11 εr + 1 + 0.23 + 84.8528 εr + 1 εr

(A2.31)

B=

592.2 √ Z0 εr

(A2.32)

and

Experimental verification indicates that (A2.29) and (A2.30) are fairly accurate as long as t/ h ≤ 0.005.

APPENDIX 3 SPECIFICATIONS OF SELECTED COAXIAL LINES AND WAVEGUIDES

TABLE A3.1 Cable Type RG(−) 8/U 9/U 11/U 58/U 59/U 122/U 141A/U 142B/U 174/U 178B/U 179B/U 180B/U 213/U 214/U 223/U 316/U

Selected Computer, Instrumentation, and Broadcast Cables

Insulation

Core O.D. (in.)

Cable O.D. (in.)

Zo ()

Capacitance (pF/ft)

Attenuation (dB/100 ft) at 400 MHz

Polyethylene Polyethylene Polyethylene Polyethylene Polyethylene Polyethylene Teflon Teflon Teflon Teflon Teflon Teflon Polyethylene Polyethylene Polyethylene Teflon

0.285 0.280 0.285 0.116 0.146 0.096 0.116 0.116 0.060 0.034 0.063 0.102 0.285 0.285 0.116 0.060

0.405 0.420 0.405 0.195 0.242 0.160 0.190 0.195 0.100 0.072 0.100 0.140 0.405 0.425 0.206 0.098

50 51 75 53.5 75 50 50 50 50 50 75 95 50 50 50 50

26.0 30.0 20.5 28.5 17.3 30.8 29.0 29.0 30.8 29.0 19.5 15.0 30.8 30.8 30.8 29.0

3.8 4.1 4.2 9.5 5.6 16.5 9.0 9.0 20.0 28.0 21.0 17.0 4.7 4.7 10.0 20.0

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SPECIFICATIONS OF SELECTED COAXIAL LINES AND WAVEGUIDES

TABLE A3.2

MIL-C-17F Designation

Selected Semirigid Coaxial Lines Nominal Impedance ()

Dielectric Diameter (in.)

Center Conductor Diameter (in.)

Maximum Operating Frequency (GHz)

Capacitance (pF/ft)

Attenuation at 1 GHz (dB/100 ft)

50 50 50 50 50

0.209 0.1175 0.066 0.037 0.026

0.0641 0.0362 0.0201 0.0113 0.008

18 20 20 20 20

29.6 29.9 32 32 32

7.5 12 22 40 60

129 130 133 151 154

TABLE A3.3

Standard Rectangular Waveguides

EIA Nomenclature WR (−)

Inside Dimension (in.)

TE10 Mode Cutoff Frequency (GHz)

Recommended Frequency Band for TE10 Mode

2300 2100 1800 1500 1150 975 770 650 510 430 340 284 229 187 159 137 112 90 75 62 51 42 34 28 22 19 15 12 10 8 7 5 4 3

23.0 × 11.5 21.0 × 10.5 18.0 × 9.0 15.0 × 7.5 11.5 × 5.75 9.75 × 4.875 7.7 × 3.85 6.5 × 3.25 5.1 × 2.55 4.3 × 2.15 3.4 × 1.7 2.84 × 1.34 2.29 × 1.145 1.87 × 0.872 1.59 × 0.795 1.372 × 0.622 1.122 × 0.497 0.9 × 0.4 0.75 × 0.375 0.622 × 0.311 0.51 × 0.255 0.42 × 0.17 0.34 × 0.17 0.28 × 0.14 0.244 × 0.112 0.188 × 0.094 0.148 × 0.074 0.122 × 0.061 0.1 × 0.05 0.08 × 0.04 0.065 × 0.0325 0.051 × 0.0255 0.043 × 0.0215 0.034 × 0.017

0.2565046 0.2809343 0.3277583 0.3933131 0.5130267 0.6051054 0.7662235 0.9077035 1.1569429 1.3722704 1.7357340 2.0782336 2.5779246 3.1530286 3.7125356 4.3041025 5.2660611 6.5705860 7.8899412 9.4951201 11.586691 14.088529 17.415732 21.184834 26.461666 31.595916 40.058509 48.54910 59.35075 74.44066 91.22728 116.47552 137.93866 174.43849

0.32–0.49 0.35–0.53 0.41–0.62 0.49–0.75 0.64–0.98 0.76–1.15 0.96–1.46 1.14–1.73 1.45–2.2 1.72–2.61 2.17–3.30 2.60–3.95 3.22–4.90 3.94–5.99 4.64–7.05 5.38–8.17 6.57–9.99 8.20–12.5 9.84–15.0 11.9–18.0 14.5–22.0 17.6–26.7 21.7–33.0 26.4–40.0 32.9–50.1 39.2–59.6 49.8–75.8 60.5–91.9 73.8–112 92.2–140 114–173 145–220 172–261 217–330

APPENDIX 4 SOME MATHEMATICAL FORMULAS

sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B tan A + tan B tan(A + B) = 1 − tan A tan B tan A − tan B tan(A − B) = 1 + tan A tan B sin2 A + cos2 A = 1 tan2 A + 1 = sec2 A cot2 A + 1 = csc2 A A−B A+B cos 2 2 A−B A+B sin sin A − sin B = 2 cos 2 2 A+B A−B cos A + cos B = 2 cos cos 2 2 sin A + sin B = 2 sin

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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SOME MATHEMATICAL FORMULAS

A+B B−A sin 2 2 2 sin A cos B = sin(A + B) + sin(A − B) cos A − cos B = 2 sin

2 cos A sin B = sin(A + B) − sin(A − B) 2 cos A cos B = cos(A + B) + cos(A − B) 2 sin A sin B = cos(A − B) − cos(A + B) 1 − cos A A sin = ± 2 2 sin 2A = 2 sin A cos A cos 2A = cos2 A − sin2 A = 2 cos2 A − 1 = 1 − 2 sin2 A 2 tan A tan 2A = 1 − tan2 A A 1 + cos A cos = ± 2 2 A 1 − cos A sin A 1 − cos A tan = ± = = 2 1 + cos A 1 + cos A sin A ej A = cos A + j sin A e−j A = cos A − j sin A sin A =

ej A − e−j A A3 A5 A7 =A− + − + ··· 2j 3! 5! 7!

ej A + e−j A A2 A4 A6 =1− + − + ··· 2 2! 4! 6! A5 A7 A3 e−j A − ej A + 2 + 17 +··· = A + tan A = j j A e + e−j A 3 15 315 cos A =

eA − e−A 2 A e + e−A cosh A = 2 sinh A eA − e−A tanh A = = A cos A e + e−A sinh A =

cosh A eA + e−A = A sinh A e − e−A sinh(A + B) = sinh A cosh B + cosh A sinh B coth A =

cosh(A + B) = cosh A cosh B + sinh A sinh B

591

592

SOME MATHEMATICAL FORMULAS

sinh(A − B) = sinh A cosh B − cosh A sinh B cosh(A − B) = cosh A cosh B − sinh A sinh B tanh A + tanh B tanh(A + B) = 1 + tanh A tanh B tanh A − tanh B tanh(A − B) = 1 − tanh A tanh B cos j A = cosh A sin j A = j sinh A cosh2 A − sinh2 A = 1 sinh j A = j sin A cosh j A = cos A tanh j A = j tan A A3 A4 A2 + + +··· 2! 3! 4 A2 A3 A4 e−A = 1 − A + − + − ··· 2 3 4! A3 A5 A7 eA − e−A =A+ + + + ··· sinh A = 2 3! 5! 7! eA + e−A A2 A4 A6 cosh A = =1+ + + +··· 2 2! 4! 6! loga xy = loga x + loga y x loga = loga x − loga y y eA = 1 + A +

loga x y = y loga x loga x = logb x × loga b =

logb x logb a

ln x = log10 x × ln 10 = 2.302585 × log10 x log10 x = ln x × log10 e = 0.434294 × ln x e = 2.718281828

APPENDIX 5 VECTOR IDENTITIES

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Až(B × C) = Bž(C × A) = Cž(A × B) A×(B × C) = B(AžC) − C(AžB) ∇(φ1 φ2 ) = φ1 ∇φ2 + φ2 ∇φ1 ∇ ž(A × B) = Bž(∇ × A) − Až(∇ × B) ∇×φA = ∇φ×A + φ(∇ × A) ∇ žφA = ∇φžA + φ(∇ žA) ∇ ž(∇ × A) = 0 ∇×(∇φ) = 0 ∇ × ∇ × A = ∇(∇ žA) − ∇ 2 A Stokes’s Theorem. The circulation around a simple closed curve is equal to the integral over any simple surface spanning the curve, of the normal component of the curl, the positive sense on the curve being counterclockwise as seen from the side of the surface toward which the positive normal points. Mathematically, Aždl = (∇ × A)žds C

S

11. Divergence Theorem (also known as Gauss’s theorem). The integral of the divergence of a vector field over a region of space is equal to the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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594

VECTOR IDENTITIES

integral over the surface of that region of the component of the field in the direction of the outward directed normal to the surface. Mathematically, ž (∇ A)dv = Ažds V

S

12. Helmholtz’s Theorem. A vector is specified uniquely if its divergence and curl are given within a region and its normal component is given over the boundary. Cartesian Coordinates ∇f = xˆ ∇2 f =

∂f ∂f ∂f + yˆ + zˆ ∂x ∂y ∂z

∂ 2f ∂ 2f ∂ 2f + 2 + 2 2 ∂x ∂y ∂z

∂Ax ∂Ay ∂Az + + ∂x ∂y ∂z xˆ yˆ zˆ ∂ ∂ ∂ ∇ × A = ∂x ∂y ∂z Ax Ay Az ∇ žA =

Cylindrical Coordinates ∂f 1 ∂f ∂f + φˆ + zˆ ∂ρ ρ ∂φ ∂z 1 ∂ ∂ 2f ∂f 1 ∂ 2f ∇2f = ρ + 2 2 + 2 ρ ∂ρ ∂ρ ρ ∂φ ∂z ∇f = ρˆ

1 ∂(ρAρ ) + ρ ∂ρ ρˆ ρφˆ ∂ ∂ ∇ × A = ∂ρ ∂φ Aρ ρAφ ∇ žA =

1 ∂Aφ ∂Az + ρ ∂φ ∂z zˆ ∂ ∂z Az

Spherical Coordinates ∂f 1 ∂f 1 ∂f + θˆ + φˆ ∂r r ∂θ r sin θ ∂φ 1 ∂ ∂ ∂f 1 ∂ 2f 2 2 ∂f ∇ f = 2 sin θ r + sin θ + r sin θ ∂r ∂r ∂θ ∂θ sin θ ∂φ2 ∇f = rˆ

VECTOR IDENTITIES

1 ∂ ∂Aφ ∂ 2 ∇A= 2 sin θ (r Ar ) + r (sin θAθ ) + r r sin θ ∂r ∂θ ∂φ rˆ r θˆ r sin θφˆ 1 ∂ ∂ ∂ ∇×A= 2 r sin θ ∂r ∂θ ∂φ Ar rAθ r sin θAφ ž

595

APPENDIX 6 SOME USEFUL NETWORK TRANSFORMATIONS

This appendix contains transformations of networks that are commonly encountered. These transformations can save time of analysis and design. Figure A6.1 shows a series-connected Rs –Ls circuit transformed to a parallel-connected Rp –Lp circuit. Resistors in the two circuits may represent loss associated with the inductor. These two circuits are equivalent if their impedance (as well as their admittance) values are equal. This equality gives Rp = Rs +

(ωLs )2 Rs

(A6.1)

Lp = Ls +

Rs2 ω2 Ls

(A6.2)

and

Transforming the other way gives Rs =

Rp (ωLp )2 Rp2 + (ωLp )2

(A6.3)

Ls =

Lp (Rp )2 Rp2 + (ωLp )2

(A6.4)

and

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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SOME USEFUL NETWORK TRANSFORMATIONS

597

Ls Lp

Rp

Rs

Figure A6.1 Transformation of a series R–L to a shunt R–L circuit, or vice versa.

Cs Cp

Rs

Figure A6.2

Rp

Transformation of a series R–C to a shunt R–C circuit, or vice versa.

Similarly, a series Rs –Cs circuit can be transformed to a parallel Rp –Cp circuit, or vice versa, as illustrated in Figure A6.2. By equating their impedance (or admittance) expression, following relations may be found: Rp =

1 + (ωCs Rs )2 Rs (ωCs )2

(A6.5)

Cp =

Cs 1 + (ωCs Rs )2

(A6.6)

Rs =

Rp 1 + (ωCp Rp )2

(A6.7)

Cs =

1 + (ωCp Rp )2 ω2 Cp Rp2

(A6.8)

and

Transforming other way gives

and

The inductor-tapped circuit shown in Figure A6.3 may be transformed to a parallel Rp –Lp circuit. This type of transformation facilitates the analysis of a Hartley oscillator. Again, the impedance of an inductor-tapped circuit must be equal to that of a parallel Rp –Lp circuit. Therefore, Rp =

(ωL1 L2 )2 + R 2 (L1 + L2 )2 RL22

(A6.9)

Lp =

(ωL1 L2 )2 + R 2 (L1 + L2 )2 ω2 L1 L22 + R 2 (L1 + L2 )

(A6.10)

and

598

SOME USEFUL NETWORK TRANSFORMATIONS

L1 Lp

L2

Rp

R

Figure A6.3

Transformation of an inductor-tapped circuit.

C1 Cp

C2

Rp

R

Figure A6.4

Transformation of a capacitor-tapped circuit.

Similarly, a capacitor-tapped circuit can be transformed to a shunt Rp –Cp circuit as illustrated in Figure A6.4. In this case, the following relations must hold if the two circuits are equal: Rp =

1 + ω2 R 2 (C1 + C2 )2 ω2 RC12

(A6.11)

Cp =

C1 [1 + ω2 R 2 C2 (C1 + C2 )] 1 + ω2 R 2 (C1 + C2 )2

(A6.12)

and

APPENDIX 7 PROPERTIES OF SOME MATERIALS

TABLE A7.1 Conductivity of Some Materials Material Aluminum Brass Bronze Copper Gold Iron Nichrome Nickel Silver Stainless steel

Conductivity (S/m) 3.54 1.5 1.0 5.813 4.1 1.04 0.09 1.15 6.12 0.11

× × × × × × × × × ×

107 107 107 107 107 107 107 107 107 107

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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600

PROPERTIES OF SOME MATERIALS

TABLE A7.2 Dielectric Constant and Loss Tangent of Some Materials at 3 GHz Material Alumina Glass (Pyrex) Mica (ruby) Nylon (610) Polystyrene Plexiglas Quartz (fused) Rexolite (1422) Styrofoam (103.7) Teflon Water (distilled)

Dielectric Constant, εr

Loss Tangent, tan δ

9.6 4.82 5.4 2.84 2.55 2.60 3.8 2.54 1.03 2.1 77

0.0001 0.0054 0.0003 0.012 0.0003 0.0057 0.00006 0.00048 0.0001 0.00015 0.157

APPENDIX 8 COMMON ABBREVIATIONS

AAU ABCMOS ABR ACI ACTS ADC ADPCM ADSL AFC AGC AM AMPS ANSI ARDIS ASCII ASIC ASK ATM AWGN

analog audio advanced bipolar CMOS available bit rate adjacent channel interference advanced communication technology satellite analog-to-digital converter adaptive differential pulse code modulation asymmetric digital subscriber line automatic frequency control automatic gain control amplitude modulation advanced mobile phone service American National Standards Institute advanced radio data information service American Standard Code for Information Interchange application-specific integrated circuit amplitude-shift keying asynchronous transfer mode additive white Gaussian noise

B-CDMA BER BIFET

broadcast CDMA bit error rate bipolar field-effect transistor

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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602

BIMOS BIOS BJT BONDING BPSK

COMMON ABBREVIATIONS

bipolar metal-oxide semiconductor basic input–output system bipolar junction transistor bandwidth on demand interoperability biphase-shift keying

CAD computer-aided design CBR continuous bit rate CCD charge-coupled device CCIR International Radio Consultative Committee CDMA code-division multiple access CDPD cellular digital packet data CIR carrier-over-interference ratio CMOS complementary metal-oxide semiconductor CNR carrier-over-noise ratio COB chip on board Codec coder–decoder CRC cyclic redundancy check CSDN circuit-switched digital network CSMA/CD carrier sense multiple access with collision detection CT cordless telephone CW continuous wave DAB DAC DAS DBS DCE DCP DCPSK DCS-1800 DDS DECT DFT DIP DMR DPCM DPSK DQPSK DRAM DRO DSMA DSO DSP DUT DWT

digital audio broadcast digital-to-analog converter data acquisition system direct broadcast satellite data communication equipment data communication protocol differentially coherent phase-shift keying digital communication system 1800 direct digital synthesis digital European cordless telecommunications discrete Fourier transforms dual-in-line package digital mobile radio differential pulse code modulation differential phase-shift keying differential quadrature phase-shift keying dynamic random access memory dielectric resonator oscillator digital sensed multiple access digital sampling oscilloscope digital signal processing device under test discrete wavelet transforms

COMMON ABBREVIATIONS

603

EBS EDA EIA EIRP EMC EMI ERC ESD ESDI ETN ETSI

emergency broadcast system electronic design automation Electronics Industries Association effective isotropic radiated power electromagnetic compatibility electromagnetic interference European Radio Commission electrostatic discharge enhanced small device interface electronic tandem network European Telecommunication Standard Institute

FCC FDD FDDI FDMA FET FFT FH FIFO FIR FM FPGA FSK

Federal Communications Commission frequency-division duplex fiber-distributed data interface frequency-division multiple access field-effect transistor fast Fourier transforms frequency hopping first-in first-out system finite impulse response filter frequency modulation field-programmable gate array frequency-shift keying

GEO GFSK GIGO GMSK GPIB GPS GSM GUI

geosynchronous satellite Gaussian frequency shift keying garbage in, garbage out Gaussian minimum shift keying general-purpose interface bus global positioning system (Groupe Special Mobile) global system for mobile communication graphical user interface

HBT HDTV HEMT HF HIPERLAN

heterojunction bipolar transistor high-definition television high-electron mobility transistor high frequency high-performance radio LAN

I and Q IC IDE IF IGBT IIT

in-phase and quadrature phase integrated circuits integrated device electronics intermediate frequency insulated gate bipolar transistor infinite impulse response filter

604

COMMON ABBREVIATIONS

IMD IMPATT INTELSAT IP IP IP2 IP3 ISDN ISI ISM ITU

intermodulation distortion impact ionization avalanche transit time diode International Telecommunication Satellite Consortium intermodulation product Internet protocol second-order intercept third-order intercept integrated services digital network intersymbol interference industrial, scientific, and medical bands International Telecommunication Union

JDC JFET JPEG

Japanese digital cellular standard junction field-effect transistor Joint Photographic Experts Group

LAN LATA LCD LED LEOs LHCP LNA LO LOS LUT

local area network local access and transport area liquid-crystal display light-emitting diode low-Earth-orbit satellite left-hand circular polarization low-noise amplifier local oscillator line of sight lookup table

MAC MAN MAU MBE MDS MDSL MEOs MESFET MIDI MIPS MLC MMI MMIC MMSI MODFET MOSFET MPEG MPP MPSK

medium access control metropolitan area network multistation access unit molecular beam epitaxy minimum detectable signal medium-bit-rate digital subscriber line medium-Earth-orbit satellite metal-semiconductor field-effect transistor musical instrument digital interface million instructions per second multilayer ceramics man–machine interface monolithic microwave integrated circuit multimode stereoscopic imaging modulation doped field-effect transistor metal-oxide-semiconductor field-effect transistor Motion Picture Experts Group massively parallel processing minimum phase-shift keying

COMMON ABBREVIATIONS

MSK MSP MSS MTI MTSO MUX

minimum shift keying mixed signal processor mobile satellite service moving-target indicator mobile telephone switching office multiplexing

NADC NF NMT NTSC NTT

North American Digital Cellular/Cordless noise figure Nordic Mobile Telephone National Television Standards Committee Nippon Telephone and Telegraph

OFDM OOK OOP

orthogonal frequency-division multiplexing on–off keying object-oriented programming

PAD PAL PAM PAMR PBX PCB PCM PCMCIA PCN PCS PDA PDC PDIP PGA PHP PHS PID PIN PLC PLL PM PMR POTS PPGA PQFP PSK PSTN PVC

packet assembler and disassembler phase alternating line pulse amplitude modulation public access mobile radio private branch exchange printed circuit board pulse code modulation Personal Computer Memory Card International Association personal communication network personal communication service personal digital assistant personal digital cellular plastic dual-in-line package programmable gain amplifier; also pin grid array personal handy phone personal handyphone system (formerly PHP) proportional integrating differentiating p-type, intrinsic, n-type diode programmable logic controller phase-locked loop pulse modulation private mobile radio plain old telephone service plastic pin grid array plastic quad flat pack phase-shift keying public switched telephone network permanent virtual connection

605

606

COMMON ABBREVIATIONS

PWB PWM

printed wire board pulse width modulation

QAM QCELP QDM QMFSK QPSK

quadrature amplitude modulation Qualcomm coded excited linear predictive coding quadrature demodulator quadrature modulation frequency-shift keying quadrature phase-shift keying

RADAR RAID RBDS RCS RES RF RFI RHCP RMS RPE-LTP RTD RTMS

radio detection and ranging redundant array of independent disks radio broadcast data system radar cross section Radio Expert Systems Group radio frequency radio-frequency interference right-handed circular polarization root mean square regular pulse excitation long-term predictor resistive thermal detector radio telephone mobile system

SAW SCSI SDH SDIP SDN SFDR SINAD SIP SLICs SMD SMR SNR SOI SOIC SONET SOT SPLIC SQPSK SQUID SSB STP SVD SWR

surface acoustic wave small computer systems interface synchronous digital hierarchy shrink dual-in-line package switched digital network spur-free dynamic range signal-over-noise and distortion single-in-line package standard linear integrated circuits surface-mount device specialized mobile radio signal-to-noise ratio silicon on insulator small outline integrated circuit synchronous optical network small outline transistor special-purpose linear integrated circuit staggered quadriphase-shift keying superconducting quantum interface single sideband shielded twisted pair simultaneous voice and data standing wave ratio

COMMON ABBREVIATIONS

607

TACS TCP T-DAB TDD TDMA TDR TE TEM TETRA THD+N TIA TM TNC TSAPI TTL TVG TVRO TVS TWTA Tx

total access communication system transmission control protocol terrestrial-digital audio broadcasting time-division duplex time-division multiple access time-domain reflectometry transverse electric transverse electromagnetic wave trans-European trunked radio system total harmonic distortion plus noise Telecommunications Industry Association transverse magnetic threaded Neil-Councilman connector telephony services–oriented application programming interface transistor–transistor logic time-variable gain TV receive-only transient voltage suppression traveling wave tube amplifier transmitter

UART UDPC UHF UNC UNI UPS USART USB UTC UTP

universal asynchronous receiver/transmitter universal digital personal communications ultrahigh frequency universal control network user-to-network interface uninterruptible power supply universal synchronous–asynchronous receiver transmitter universal serial bus universal time code unshielded twisted pair

VCO VDT VHDL VHF VLF VLSI VPSK VSAT VSELP VSWR VXCO

voltage-controlled oscillator video display terminal very high definition language very high frequency very low frequency very-large-scale integration variable phase-shift keying very small aperture (satellite ground) terminal vector sum excited linear predictive coding voltage standing wave ratio voltage-controlled crystal oscillator

WAN WATS

wide area network wide area telecommunication network

608

WDM WER WFAU WLAN WLL WORM WPABX

COMMON ABBREVIATIONS

wavelength-division multiplexing word error rate wireless fixed access unit wireless local area network wireless local loop write once, read many wireless private branch exchange

APPENDIX 9 PHYSICAL CONSTANTS

Permittivity of free space, ε0 Permeability of free space, µ0 Impedance of free space, η0 Velocity of light in free space, c Charge of electron, qe Mass of electron, me Boltzmann’s constant, k Planck’s constant, h

8.8542 × 10−12 F/m 4π × 10−7 H/m 376.7 2.997925 × 108 m/s 1.60210 × 10−19 C 9.1091 × 10−31 kg 1.38 × 10−23 J/K 6.6256 × 10−34 J·s

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

609

INDEX

1/f noise, 34 Acquisition of lock, 509 Admittance matrix, 289 Admittance parameters, 289 AM, see Amplitude modulation Ampere’s law, 105 Amplitude modulation, 544 DSB, 546 modulation index, 544 SSB, 546 suppressed carrier AM, 546 Antenna efficiency, 21 Antennas, 18 Cassegrain reflector, 18 directive gain, 19 electric dipole, 18 isotropic antenna, 19 lens, 18 power gain, 19 pyramidal horn, 18 Attenuation constant, 62, 116 Available power gain, 411 Balanced amplifier, 468 Bandpass filter, 371 Bandstop filter, 375 Barkhausen criterion, 480

Bessel function, 143, 180, 556 Binomial filter, 354 Binomial transformers, 244 Bisected π-section, 346 Black-body radiation law, 34 Bode–Fano constraints, 280 Broadband amplifiers, 466 Butterworth filter, 354 C3 , 7 Carson rule bandwidth, 556 Cellular telephones, 17 Chain matrix, 298 Chain scattering parameters, 325 Characteristic impedance, 59 experimental determination of, 75 Chebyshev filter, 355 Chebyshev polynomials, 250, 356 Chebyshev transformers, 248, 255 Clapp oscillator, 488 Closed-loop gain, 480 Coaxial line, 582 Coefficient of coupling, 166 Coherent detector, 553 Colpitts oscillator, 484 Complex permittivity, 112 Composite filters, 346

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

611

612 Conductivity, 59, 112 Conservation of charge, 107 Constant gain circles, 439 Constant noise figure circles, 457 Constant-k, 337 Constitutive relations, 111 Cordless telephones, 17 Critically damped, 153 Current reflection coefficient, 73 Current sensitivity, 551 Cutoff frequency, 337, 338, 354, 360, 369 Cylindrical cavity, 179 Cylindrical waveguide, 137, 142 Damping ratio, 153 dB, 580 dBc, 581 dBW, 580 Dibit, 9 Directivity, 19, 27 Distortion, 33, 45 Distortionless line, 75 Distributed amplifiers, 468 Distributed elements, 58 Doppler radar, 11, 32 Divergence theorem, 593 Double-balanced mixer, 563 Double-stub tuner, 202 DR, see Dielectric resonators Dynamic conductance, 550 Dynamic range, 5, 11, 47 EIRP, 24 Electric charge density, 105 Electric current density, 105 Electric field intensity, 104, 112 Electric flux density, 105, 111 Electric scalar potential, 128 Electric vector potential, 129 Electrical susceptibility, 111 Equation of continuity, 107 Faraday’s law of induction, 105 FDM, 543 FET mixer, 571 Flicker noise, 34 FM, see Frequency modulation FM detector, 558 Fractional bandwidth, 236 Frequency bands, 2, 3 Frequency converter, 547 Frequency division multiplexing, 543 Frequency modulation, 2, 9, 543, 555 modulation index, 555

INDEX Frequency scaling, 360 Frequency-shift keying, 9 Friis transmission formula, 26 Gain compression, 48 Geosynchronous, 13 Gilbert cell, 575 Global positioning system, 15 Group velocity, 65 Gunn diodes, 520 Half-power beam width, 20 Hankel functions, 144 Hartley oscillator, 484 HBT, 422 Helmholtz’s equations, 62 Helmholtz’s theorem, 128 HEMT, 422 High-pass filter, 338, 369 Hybrid parameters, 296 Image impedance, 334 Image parameter method, 334 IMD, 569 IMPATT diode, 520 Impedance matrix, 284 Impedance parameters, 284 Impedance scaling, 361 Incident wave, 64, 123 Indirect frequency synthesizer, 516 Input node, 396 Input reflection coefficient, 68 Input stability circle, 427 Insertion loss, 72 Insertion loss method, 353 Intercept point, 50 Intermediate frequency, 12 Intermodulation distortion, 45, 567 Intermodulation distortion ratio, 49 Intrinsic impedance, 115 IP, see Intercept point IR, 2 Isotropic antenna, 19 Johnson noise, 33 Klopfenstein taper, 276 Kuroda’s identities, 384 Leontovich impedance boundary condition, 121 LEOS, 15, 16 Line-of-sight propagation, 5 Load reflection coefficient, 70, 80

INDEX Lock-in range, 510 Loop, 398, 480 Loop gain, 398, 480 Loop-filters, 500 Lorentz condition, 128, 130 Loss-tangent, 112 Low-pass filter, 337, 354 Low-pass filter synthesis, 358 Low-pass ladder network prototype, 358 Magnetic charge, 126 Magnetic current, 126 Magnetic field intensity, 104 Magnetic flux density in, 105 Magnetic scalar potential, 129 Magnetic susceptibility, 112 Magnetic vector potential, 127 Magnetization density, 112 Maximally flat filter, 354 Maximally flat low-pass filters, 355 Maxwell’s equations, 109, 110 m-derived filter, 341 MEOS, 15 Microstrip line, 585 Microwave transistor oscillator, 523 Minimum detectable signal, 44 Minkowski inequality, 424 Mixer, 9, 12 Mixers: conversion loss, 565 diode ring, 567 dual gate FET, 574 FET mixers, 571 intermodulation distortion, 569 single diode, 548 switching type, 559 Negative resistance, 489 Neper, 581 Neumann function, 143 Noise, 33 Noise factor, 37 Noise figure, 38 Noise figure of a cascaded system, 40 Noise figure parameter, 458 Noise in two-port networks, 39 Noise temperature, 35 Nonredundant circuit synthesis, 380 Normalized image impedance, 344 Normalized input impedance, 68 Notch-filter, 333 Nyquist criterion, 480 Nyquist noise, 33

On-off keying, 9 Operating power gain, 414 Output node, 396 Output stability circle, 426 Overdamped, 153 Parallel-plate waveguide, 133 Path, 398 Path gain, 398 Personal communication networks, 16 Personal handy phone, 17 Phase constant, 62, 114 Phase detector, 498 Phase modulation, 9 Phase velocity, 65, 115 Phase-locked loop, 497 Phase-shift keying, 9 Pierce oscillator, 494 Pink noise, 34 Polarization, 23 Polarization density, 111 Power density, 19 Power flow, 113 Power loss ratio, 353 Poynting’s vector, 113 Propagation constant, 62, 116 experimental determination of, 75 Pull-in range, 510 Pulling figure, 494 Q, see Quality factor Quadrature phase-shift keying, 9 Quality factor, 155 Quarter-wavelength transformer, 70 Radar, 4 Radar cross-sections, 29 Radar range equation, 29 Radiation efficiency, 22 Radiation field, 19 Radiation intensity, 19 Radiation pattern, 20 Radio frequency detector, 551 Radio frequency identification, 2 Rayleigh–Jeans approximation, 34 Reactive L-section matching, 209 Receiver, 8 Rectangular cavity, 177 Redundant filter synthesis, 380 Reflected wave, 64 Reflection coefficient, 68 Remote sensing, 7 Repeater, 12 Resistive L-section matching, 207

613

614

INDEX

Resonant frequency, 179 Return-loss, 72 Reverse-saturation current, 549 Richard’s transformation, 384 Sampling theorem, 272 Satellite communication, 13 Scattering matrix, 306 Scattering parameters, 306 shifting of reference planes, 311 Scattering transfer parameters, 325 Second harmonic distortion, 48 Self-loop, 398 Semirigid coaxial lines, 589 Shot noise, 33 Shunt matching element, 190 Shunt stub matching, 190 Signal flow graph, 392 branches, 392 junction points, 392 Mason’s rule, 399 nodes, 663 rules of reduction, 398 Signal-to-noise ratio, 37 Skin depth, 115 Simultaneous conjugate matching, 432 Single-balanced mixer, 560 Single-diode mixer circuit, 548 Small-signal equivalent circuit of a BJT, 469 Small-signal equivalent circuit of a MESFET, 471 Small-signal equivalent circuit of a MOSFET, 470 Smith chart, 85 Space loss, 25 Spurious-free dynamic range, 51 SSB, 546 Stability circle, 427 Stokes’ theorem, 593 Strip line, 583 Stub, 189 Switching-type mixers, 559 Synchronization range, 507 Terrestrial communication, 12 Three-port S-parameter description of the transistor, 530 Time-harmonic fields, 110

Time-period, 64 Tracking range, 507 Transducer power gain, 412 Transmission line: attenuation constant, 62 characteristic impedance, 59 distributed network model, 61 line parameters, 58 phase constant, 62 propagation constant, 62 Transmission line equations, 61 Transmission lines, 4 Transmission parameters, 298 Transmitter, 8 Transponder, 14 Traveling wave amplifiers, 468 Traveling wave tube, 14 Tuning sensitivity, 496 UHF, 3 Undamped natural frequency, 153 Underdamped, 153 Unilateral figure of merit, 437 Unilateral transducer power gain, 413 Varactors, 495 VCO, see Voltage-controlled oscillator Vector wave equation, 114 VHF, 3 Voltage-controlled oscillator, 496, 497 Voltage reflection coefficient, 72 Voltage sensitivity, 552 Voltage standing wave, 82 VSAT, 15 VSWR, 82 Wavelength, 64 Wave number, 114 Wire antennas, 18 electric dipole, 18 loop antennas, 18 Wireless communication, 8, 16 Y -factor method, 36 Zero-biased diode detector, 552 ZY -Smith chart, 219

RADIO-FREQUENCY AND MICROWAVE COMMUNICATION CIRCUITS Analysis and Design Second Edition

Devendra K. Misra University of Wisconsin–Milwaukee

A JOHN WILEY & SONS, INC., PUBLICATION

Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax 978-646-8600, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services please contact our Customer Care Department within the U.S. at 877-762-2974, outside the U.S. at 317-572-3993 or fax 317-572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print, however, may not be available in electronic format. Library of Congress Cataloging-in-Publication Data: Misra, Devendra, 1949– Radio-frequency and microwave communication circuits : analysis and design / Devendra K. Misra.—2nd ed. p. cm. Includes bibliographical references and index. ISBN 0-471-47873-3 (Cloth) 1. Radar circuits–Design and construction. 2. Microwave circuits–Design and construction. 3. Electronic circuit design. 4. Radio frequency. I. Title. TK6560 .M54 2004 621.384 12–dc22 2003026691 Printed in the United States of America. 10 9 8 7 6 5 4 3 2 1

CONTENTS

Preface

ix

1 Introduction 1.1 Microwave Transmission Lines 1.2 Transmitter and Receiver Architectures 2 Communication Systems 2.1 2.2 2.3 2.4 2.5

Terrestrial Communication Satellite Communication Radio-Frequency Wireless Services Antenna Systems Noise and Distortion Suggested Reading Problems

3 Transmission Lines 3.1 3.2 3.3 3.4

Distributed Circuit Analysis of Transmission Lines Sending-End Impedance Standing Wave and Standing Wave Ratio Smith Chart Suggested Reading Problems

1 4 8 11 12 13 16 18 33 52 52 57 57 67 80 85 95 95 v

vi

CONTENTS

4 Electromagnetic Fields and Waves 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

Fundamental Laws of Electromagnetic Fields The Wave Equation and Uniform Plane Wave Solutions Boundary Conditions Uniform Plane Wave Incident Normally on an Interface Modified Maxwell’s Equations and Potential Functions Construction of Solutions Metallic Parallel-Plate Waveguide Metallic Rectangular Waveguide Metallic Circular Waveguide Suggested Reading Problems

5 Resonant Circuits 5.1 5.2 5.3 5.4 5.5

Series Resonant Circuits Parallel Resonant Circuits Transformer-Coupled Circuits Transmission Line Resonant Circuits Microwave Resonators Suggested Reading Problems

6 Impedance-Matching Networks 6.1 Single Reactive Element or Stub Matching Networks 6.2 Double-Stub Matching Networks 6.3 Matching Networks Using Lumped Elements Suggested Reading Problems 7 Impedance Transformers 7.1 Single-Section Quarter-Wave Transformers 7.2 Multisection Quarter-Wave Transformers 7.3 Transformer with Uniformly Distributed Section Reflection Coefficients 7.4 Binomial Transformers 7.5 Chebyshev Transformers 7.6 Exact Formulation and Design of Multisection Impedance Transformers 7.7 Tapered Transmission Lines 7.8 Synthesis of Transmission Line Tapers 7.9 Bode–Fano Constraints for Lossless Matching Networks

104 104 114 119 123 126 130 133 137 142 145 145 151 151 160 164 170 177 184 184 189 190 202 207 226 226 234 234 237 239 244 248 255 263 270 280

CONTENTS

Suggested Reading Problems 8 Two-Port Networks 8.1 8.2 8.3 8.4 8.5

Impedance Parameters Admittance Parameters Hybrid Parameters Transmission Parameters Conversion of Impedance, Admittance, Chain, and Hybrid Parameters 8.6 Scattering Parameters 8.7 Conversion From Impedance, Admittance, Chain, and Hybrid Parameters to Scattering Parameters, or Vice Versa 8.8 Chain Scattering Parameters Suggested Reading Problems 9 Filter Design 9.1 Image Parameter Method 9.2 Insertion-Loss Method 9.3 Microwave Filters Suggested Reading Problems 10 Signal-Flow Graphs and Their Applications 10.1 Definitions and Manipulation of Signal-Flow Graphs 10.2 Signal-Flow Graph Representation of a Voltage Source 10.3 Signal-Flow Graph Representation of a Passive Single-Port Device 10.4 Power Gain Equations Suggested Reading Problems 11 Transistor Amplifier Design 11.1 11.2 11.3 11.4 11.5 11.6 11.7

Stability Considerations Amplifier Design for Maximum Gain Constant-Gain Circles Constant Noise Figure Circles Broadband Amplifiers Small-Signal Equivalent-Circuit Models of Transistors DC Bias Circuits for Transistors

vii

281 281 283 284 289 296 298 304 304 323 325 325 326 333 334 353 380 389 389 392 396 401 402 410 417 418 422 422 429 439 457 466 469 472

viii

CONTENTS

Suggested Reading Problems 12 Oscillator Design 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Feedback and Basic Concepts Crystal Oscillators Electronic Tuning of Oscillators Phase-Locked Loop Frequency Synthesizers One-Port Negative Resistance Oscillators Microwave Transistor Oscillators Suggested Reading Problems

13 Detectors and Mixers 13.1 13.2 13.3 13.4 13.5 13.6

Amplitude Modulation Frequency Modulation Switching-Type Mixers Conversion Loss Intermodulation Distortion in Diode-Ring Mixers FET Mixers Suggested Reading Problems

475 476 479 479 492 494 497 516 520 523 538 538 543 544 555 559 565 567 571 577 577

Appendix 1

Decibels and Neper

580

Appendix 2

Characteristics of Selected Transmission Lines

582

Appendix 3

Specifications of Selected Coaxial Lines and Waveguides

588

Appendix 4

Some Mathematical Formulas

590

Appendix 5

Vector Identities

593

Appendix 6

Some Useful Network Transformations

596

Appendix 7

Properties of Some Materials

599

Appendix 8

Common Abbreviations

601

Appendix 9

Physical Constants

609

Index

611

PREFACE

Wireless technology continues to grow at a tremendous rate, with new applications still reported almost daily. In addition to the traditional applications in communications, such as radio and television, radio-frequency (RF) and microwaves are being used in cordless phones, cellular communication, local area networks, and personal communication systems. Keyless door entry, radio-frequency identification, monitoring of patients in a hospital or a nursing home, cordless mice or keyboards for computers, and wireless networking of home appliances are some of the other areas where RF technology is being employed. Although some of these applications have traditionally used infrared technology, RF circuits are taking over, because of their superior performance. The present rate of growth in RF technology is expected to continue in the foreseeable future. These advances require the addition of personnel in the areas of radio-frequency and microwave engineering. Therefore, in addition to regular courses as a part of electrical engineering curriculums, short courses and workshops are regularly conducted in these areas for practicing engineers. This edition of the book maintains the earlier approach of a presentation based on a basic course in electronic circuits. At the same time, a new chapter on electromagnetic fields has been added, following several constructive suggestions from those who used the first edition. It provides the added option of using the book for a traditional microwave engineering course with electromagnetic fields and waves. Or, this chapter can be bypassed so as to follow the approach used in the first edition: that is, instead of using electromagnetic fields as most microwave engineering books do, the subject is introduced via circuit concepts. Further, an overview of communication systems is presented in the beginning to provide the reader with an overall perspective of various building blocks involved. ix

x

PREFACE

This edition of the book is organized into thirteen chapters and nine appendixes, using a top-down approach. It begins with an introduction to frequency bands, RF and microwave devices, and applications in communication, radar, industrial, and biomedical areas. The introduction includes a brief description of microwave transmission lines: waveguides, strip lines, and microstrip line. An overview of transmitters and receivers is included, along with digital modulation and demodulation techniques. Modern wireless communication systems, such as terrestrial and satellite communication systems and RF wireless services, are discussed briefly in Chapter 2. After introducing antenna terminology, effective isotropic radiated power, the Friis transmission formula, and the radar range equation are presented. In the final section of the chapter, noise and distortion associated with communication systems are introduced. Chapter 3 begins with a discussion of distributed circuits and construction of a solution to the transmission line equation. Topics presented in this chapter include RF circuit analysis, phase and group velocities, sending-end impedance, reflection coefficient, return loss, insertion loss, experimental determination of characteristic impedance and the propagation constant, the voltage standing wave ratio, and impedance measurement. The final section in Chapter 3 includes a description of the Smith chart and its application in the analysis of transmission line circuits. Fundamental laws of electromagnetic fields are introduced in Chapter 4 along with wave equations and uniform plane wave solutions. Boundary conditions and potential functions that lead to the construction of solutions are then introduced. The chapter concludes with analyses of various metallic waveguides. Resonant circuits are discussed in Chapter 5, which begins with series and parallel resistance–inductance–capacitance circuits. This is followed by a section on transformer-coupled circuits. The final two sections of the chapter are devoted to transmission line resonant circuits and microwave resonators. Chapters 6 and 7 deal with impedance-matching techniques. Single reactive element or stub, double-stub, and lumped-element matching techniques are discussed in Chapter 6. Chapter 7 is devoted to multisection transmission line impedance transformers, binomial and Chebyshev sections, and impedance tapers. Chapter 8 introduces circuit parameters associated with two-port networks. Impedance, admittance, hybrid, transmission, scattering, and chain scattering parameters are presented, along with examples that illustrate their characteristic behaviors. Chapter 9 begins with the image parameter method for the design of passive filter circuits. The insertion-loss technique is introduced next to synthesize Butterworth and Chebyshev low-pass filters. The chapter includes descriptions of impedance and frequency scaling techniques to realize high-pass, bandpass, and bandstop networks. The chapter concludes with a section on microwave transmission line filter design. Concepts of signal-flow-graph analysis are introduced in Chapter 10, along with a representation of voltage source and passive devices, which facilitates formulation of the power gain relations that are needed in the amplifier design discussed in Chapter 11. The chapter begins with stability considerations using

PREFACE

xi

scattering parameters of a two-port network followed by design techniques of various amplifiers. Chapter 12 presents basic concepts and design of various oscillator circuits. The phase-locked loop and its application in the design of frequency synthesizers are also summarized. The final section of the chapter includes the analysis and design of microwave transistor oscillators using S-parameters. Chapter 13 includes the fundamentals of frequency-division multiplexing, amplitude modulation, radio-frequency detection, frequency-modulated signals, and mixer circuits. The book ends with nine appendixes, which include a discussion of logarithmic units (dB, dBm, dBW, dBc, and neper), design equations for selected transmission lines (coaxial line, strip line, and microstrip line), and a list of abbreviations used in the communication area. Some of the highlights of the book are as follows:

ž

The presentation begins with an overview of frequency bands, RF and microwave devices, and their applications in various areas. Communication systems are presented in Chapter 2, including terrestrial and satellite systems, wireless services, antenna terminology, the Friis transmission formula, the radar equation, and Doppler radar. Thus, students learn about the systems using blocks of amplifiers, oscillators, mixers, filters, and so on. Students’ response has strongly supported this top-down approach.

ž

Since students are assumed to have only one semester of electrical circuits, resonant circuits and two-port networks are included in the book. Concepts of network parameters (impedance, admittance, hybrid, transmission, and scattering) and their characteristics are introduced via examples.

ž

A separate chapter on oscillator design includes concepts of feedback, the Hartley oscillator, the Colpitts oscillator, the Clapp oscillator, crystal oscillators, phased-locked-loop and frequency synthesizers, transistor oscillator design using S-parameters, and three-port S-parameter description of transistors and their use in feedback network design.

ž

A separate chapter on detectors and mixers includes amplitude- and frequency-modulated signal characteristics and their detection schemes, single-diode mixers, RF detectors, double-balanced mixers, conversion loss, intermodulation distortion in diode-ring mixers, and field-effecttransistor mixers.

ž

Appendixes include logarithmic units, design equations for selected transmission lines, and a list of abbreviations used in the communication area.

ž

There are 153 solved examples with a step-by-step explanation. Therefore, practicing engineers will find the book useful for self-study as well.

ž

There are 275 class-tested problems at the ends of chapters. Supplementary material is available to instructors adopting the book.

xii

PREFACE

Acknowledgments I learned this subject from engineers and authors who are too many to include in this short space, but I gratefully acknowledge their contributions. I would like to thank my anonymous reviewers, instructors here and abroad who used the first edition of this book and provided a number of constructive suggestions, and my former students, who made useful suggestions to improve the presentation. I deeply appreciate the support I received from my wife, Ila, and son, Shashank, during the course of this project. The first edition of this book became a reality only because of enthusiastic support from then-senior editor Philip Meyler and his staff at Wiley. I feel fortunate to continue getting the same kind of support from current editor Val Moliere and from Kirsten Rohstedt. DEVENDRA K. MISRA

1 INTRODUCTION

Scientists and mathematicians of the nineteenth century laid the foundation of telecommunication and wireless technology, which has affected all facets of modern society. In 1864, James C. Maxwell put forth fundamental relations of electromagnetic fields that not only summed up the research findings of Laplace, Poisson, Faraday, Gauss, and others but also predicted the propagation of electrical signals through space. Heinrich Hertz subsequently verified this in 1887, and Guglielmo Marconi transmitted wireless signals across the Atlantic Ocean successfully in 1900. Interested readers may find an excellent discussion of the historical developments of radio frequencies (RFs) and microwaves in the IEEE Transactions on Microwave Theory and Technique (Vol. MTT-32, September 1984). Wireless communication systems require high-frequency signals for the efficient transmission of information. Several factors lead to this requirement. For example, an antenna radiates efficiently if its size is comparable to the signal wavelength. Since the signal frequency is inversely related to its wavelength, antennas operating at RFs and microwaves have higher radiation efficiencies. Further, their size is relatively small and hence convenient for mobile communication. Another factor that favors RFs and microwaves is that the transmission of broadband information signals requires a high-frequency carrier signal. In the case of a single audio channel, the information bandwidth is about 20 kHz. If amplitude modulation (AM) is used to superimpose this information on a carrier, it requires at least this much bandwidth on one side of the spectrum. Further, Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

1

2

INTRODUCTION

TABLE 1.1 Frequency Bands Used in Commercial Broadcasting

AM TV FM TV

Channels

Frequency Range

Wavelength Range

107 2–4 5–6 100 7–13 14–83

535–1605 kHz 54–72 MHz 76–88 MHz 88–108 MHz 174–216 MHz 470–890 MHz

186.92–560.75 m 4.17–5.56 m 3.41–3.95 m 2.78–3.41 m 1.39–1.72 m 33.7–63.83 cm

commercial AM transmission requires a separation of 10-kHz between the two transmitters. On the other hand, the required bandwidth increases significantly if frequency modulation (FM) is used. Each FM transmitter typically needs a bandwidth of 200 kHz for audio transmission. Similarly, each television channel requires about 6 MHz of bandwidth to carry the video information as well. Table 1.1 shows the frequency bands used for commercial radio and television broadcasts. In the case of digital transmission, a standard monochrome television picture is sampled over a grid of 512 × 480 elements called pixels. Eight bits are required to represent 256 shades of the gray display. To display motion, 30 frames are sampled per second; thus, it requires about 59 Mb/s (512 × 480 × 8 × 30 = 58,982,400). Color transmission requires even higher bandwidth (on the order of 90 Mb/s). Wireless technology has been expanding very fast, with new applications reported every day. In addition to the traditional applications in communication, such as radio and television, RF and microwave signals are being used in cordless phones, cellular communication, local, wide, and metropolitan area networks and personal communication service. Keyless door entry, radio-frequency identification (RFID), monitoring of patients in a hospital or a nursing home, and cordless mice or keyboards for computers are some of the other areas where RF technology is being used. Although some of these applications have traditionally used infrared (IR) technology, current trends favor RF, because RF is superior to infrared technology in many ways. Unlike RF, infrared technology requires unobstructed line-of-sight connection. Although RF devices have been more expensive than IR, the current trend is downward because of an increase in their production and use. The electromagnetic frequency spectrum is divided into bands as shown in Table 1.2. Hence, AM radio transmission operates in the medium-frequency (MF) band, television channels 2 to 12 operate in the very high frequency (VHF) band, and channels 18 to 90 operate in the ultrahigh-frequency (UHF) band. Table 1.3 shows the band designations in the microwave frequency range. In addition to natural and human-made changes, electrical characteristics of the atmosphere affect the propagation of electrical signals. Figure 1.1 shows

3

INTRODUCTION

TABLE 1.2

IEEE Frequency Band Designations

Band Designation VLF LF MF HF VHF UHF SHF EHF

TABLE 1.3

Frequency Range 3–30 kHz 30–300 kHz 300–3000 kHz 3–30 MHz 30–300 MHz 300–3000 MHz 3–30 GHz 30–300 GHz

Wavelength Range (in Free Space) 10–100 km 1–10 km 100 m–1 km 10–100 m 1–10 m 10 cm–1 m 1–10 cm 0.1–1 cm

Microwave Frequency Band Designations

Frequency Bands 500–1000 MHz 1–2 GHz 2–4 GHz 3–4 GHz 4–6 GHz 6–8 GHz 8–10 GHz 10–12.4 GHz 12.4–18 GHz 18–20 GHz 20–26.5 GHz 26.5–40 GHz

Old (Still Widely Used)

New (Not So Commonly Used)

UHF L S S C C X X Ku K K Ka

C D E F G H I J J J K K

various layers of the ionosphere and the troposphere that are formed due to the ionization of atmospheric air. As illustrated in Figure 1.2(a) and (b), an RF signal can reach the receiver by propagating along the ground or after reflection from the ionosphere. These signals may be classified as ground and sky waves, respectively. The behavior of a sky wave depends on the season, day or night, and solar radiation. The ionosphere does not reflect microwaves, and the signals propagate line of sight, as shown in Figure 1.2(c). Hence, curvature of the Earth limits the range of a microwave communication link to less than 50 km. One way to increase the range is to place a human-made reflector up in the sky. This type of arrangement is called a satellite communication system. Another way to increase the range of a microwave link is to place repeaters at periodic intervals. This is known as a terrestrial communication system. Figures 1.3 and 1.4 list selected devices used at RF and microwave frequencies. Solid-state devices as well as vacuum tubes are used as active elements in RF and

4

INTRODUCTION F2 layer

F1 layer 300 km

E layer 200 km Ionosphere

125 km D layer 100 km 90 km 60 km

Troposphere 10 km

Figure 1.1

Atmosphere surrounding Earth.

microwave circuits. Predominant applications for microwave tubes are in radar, communications, electronic countermeasures (ECMs), and microwave cooking. They are also used in particle accelerators, plasma heating, material processing, and power transmission. Solid-state devices are employed primarily in the RF region and in low-power microwave circuits such as low-power transmitters for local area networks and receiver circuits. Some applications of solid-state devices are listed in Table-1.4. Figure 1.5 lists some applications of microwaves. In addition to terrestrial and satellite communications, microwaves are used in radar systems as well as in various industrial and medical applications. Civilian applications of radar include air traffic control, navigation, remote sensing, and law enforcement. Its military uses include surveillance; guidance of weapons; and command, control, and communication (C3 ). Radio-frequency and microwave energy are also used in industrial heating and for household cooking. Since this process does not use a conduction mechanism for the heat transfer, it can improve the quality of certain products significantly. For example, the hot air used in a printing press to dry ink affects the paper adversely and shortens the product’s life span. By contrast, in microwave drying only the ink portion is heated, and the paper is barely affected. Microwaves are also used in material processing, telemetry, imaging, and hyperthermia. 1.1 MICROWAVE TRANSMISSION LINES Figure 1.6 shows selected transmission lines used in RF and microwave circuits. The most common transmission line used in the RF and microwave range is the

5

MICROWAVE TRANSMISSION LINES

Receiving antenna

Transmitting antenna

Earth

(a) Signal propagation along the ground Ionosphere

Earth Receiving antenna

Transmitting antenna

(b) Propagation of a sky wave

Transmitting antenna

Earth

Receiving antenna

(c) Line-of-sight propagation

Figure 1.2

Modes of signal propagation.

coaxial line. A low-loss dielectric material is used in these transmission lines to minimize signal loss. Semirigid coaxial lines with continuous cylindrical conductors outside perform well in the microwave range. To ensure single-mode transmission, the cross section of a coaxial line must be much smaller than the signal wavelength. However, this limits the power capacity of these lines. In

6

INTRODUCTION Microwave Devices Active Solid-State

Vacuum-Tube Linear Beam Klystrons Traveling wave tubes Hybrid tubes

Passive Directional couplers Attenuators Cross-Field Phase shifters Magnetrons Antennas Resonators

Figure 1.3 Microwave devices. Solid-State Devices

Transistors

Transferred Electron

BJT FET HEMT

Gunn diode LSA diode InP diode CdTe diode

Figure 1.4 TABLE 1.4

Avalanche Transit Time BARITT diode IMPATT diode TRAPATT diode Parametric devices

Quantum Electronic Ruby masers Semiconductor laser

Solid-state devices used at RF and microwave frequencies.

Selected Applications of Microwave Solid-State Devices

Devices Transistors

Transferred electron devices (TED)

IMPATT diode

TRAPATT diode

BARITT diode

Applications

Advantages

L-band transmitters for telemetry systems and phased-array radar systems; transmitters for communication systems C-, X-, and Ku-band ECM amplifiers for wideband systems; X- and Ku-band transmitters for radar systems, such as traffic control Transmitters for millimeter-wave communication S-band pulsed transmitter for phased-array radar systems

Low cost, low power supply, reliable, high-continuouswave (CW) power output, lightweight

Local oscillators in communication and radar receivers

Low power supply (12 V), low cost, lightweight, reliable, low noise, high gain

Low power supply, low cost, reliable, high-CW power, lightweight High peak and average power, reliable, low power supply, low cost Low power supply, low cost, low noise, reliable

7

MICROWAVE TRANSMISSION LINES Microwave Applications Communication

Radar

Industrial and Biomedical

Terrestrial Civilian Air traffic Satellite control Aircraft navigation Ship safety

Military

Space vehicles Remote sensing

Heating

Surveillance

Process control Drying

Navigation

Curing

Guidance of weapons

Treatment of elastomers

Electronic warfare

Waste treatment

Industrial Household

Nuclear

C3

Cellulosic

Law enforcement

Monitoring Imaging Hyperthermia

Figure 1.5

Some applications of microwaves.

(b) Rectangular waveguide

(a) Coaxial line

(d) Ridged waveguide

(e) Fin line

εr (f) Strip line

Figure 1.6

(c) Circular waveguide

εr (g) Microstrip line

εr (h) Slot line

Transmission lines used in RF and microwave circuits.

high-power microwave circuits, waveguides are used in place of coaxial lines. Rectangular waveguides are commonly employed for connecting high-power microwave devices because these are easy to manufacture compared with circular waveguides. However, certain devices (e.g., rotary joints) require a circular cross section. In comparison with a rectangular waveguide, a ridged waveguide provides broadband operation. The fin line shown in Figure 1.6(e) is commonly

8

INTRODUCTION

used in the millimeter-wave band. Physically, it resembles a slot line enclosed in a rectangular waveguide. The transmission lines illustrated in Figure 1.6(f) to (h) are most convenient in connecting the circuit components on a printed circuit board (PCB). The physical dimensions of these transmission lines depend on the dielectric constant εr of insulating material and on the operating frequency band. The characteristics and design formulas of selected transmission lines are given in the appendixes.

1.2 TRANSMITTER AND RECEIVER ARCHITECTURES Wireless communication systems require a transmitter at one end to send the information signal and a receiver at the other to retrieve it. In one-way communication (such as a commercial broadcast), a transmitting antenna radiates the signal according to its radiation pattern. The receiver, located at the other end, receives this signal via its antenna and extracts the information, as illustrated in Figure 1.7. Thus, the transmitting station does not require a receiver, and vice versa. On the other hand, a transceiver (a transmitter and a receiver) is needed at both ends to establish a two-way communication link. Figure 1.7 is a simplified block diagram of a one-way communication link. At the transmitting end, an information signal is modulated and mixed with a local oscillator to up-convert the carrier frequency. Bandpass filters are used before

Up-converter Information signal

Modulator

BPF

BPF Power amplifier LO (a) Downconverter

Information signal

Demodulator

BPF/IF amp.

BPF

LO

Low-noise Frequencyamplifier selective circuit

(b)

Figure 1.7 Simplified block diagram of the transmitter (a) and receiver (b) of a wireless communication system.

TRANSMITTER AND RECEIVER ARCHITECTURES

9

and after the mixer to stop undesired harmonics. The signal power is amplified before feeding it to the antenna. At the receiving end the entire process is reversed to recover the information. Signal received by the antenna is filtered and amplified to improve the signal-to-noise ratio before feeding it to the mixer for down-converting the frequency. A frequency-selective amplifier (tuned amplifier) amplifies it further, before feeding it to a suitable demodulator, which extracts the information signal. In an analog communication system, the amplitude or angle (frequency or phase) of the carrier signal is varied according to the information signal. These modulations are known as amplitude modulation (AM), frequency modulation (FM), and phase modulation (PM), respectively. In a digital communication system, the input passes through channel coding, interleaving, and other processing before it is fed to the modulator. Various modulation schemes are available, including on–off keying (OOK), frequency-shift keying (FSK), and phase-shift keying (PSK). As these names suggest, a high-frequency signal is turned on and off in OOK to represent the logic states 1 and 0 of a digital signal. Similarly, two different signal frequencies are employed in FSK to represent the two signal states. In PSK, the phase of the high-frequency carrier is changed according to the 1 or 0 state of the signal. If only two phase states (0◦ and 180◦ ) of the carrier are used, it is called binary phase-shift keying (BPSK). On the other hand, a phase shift of 90◦ gives four possible states, each representing 2 bits of information (known as a dibit). This type of digital modulation, known as quadrature phase-shift keying (QPSK), is explained further below. Consider a sinusoidal signal Smod , as given by equation (1.2.1). Its angular frequency and phase are ω radians per second and φ, respectively: Smod (t) =

√

2 cos(ωt − φ) =

√

2(cos ωt cos φ + sin ωt sin φ)

(1.2.1)

This can be simplified further as follows: Smod (t) = SI cos ωt + SQ sin ωt where SI = and SQ =

√

2 cos φ

√

2 sin φ

(1.2.2)

(1.2.3)

(1.2.4)

The subscripts I and Q represent in-phase and quadrature-phase components of Smod . Table 1.5 shows the values of SI and SQ for four different phase states together with the corresponding dibit representations. This scheme can be implemented easily for a polar digital signal (positive peak value representing logic 1 and the negative peak logic 0). It is illustrated in Figure 1.8. The demultiplexer simultaneously feeds one bit of the digital input SBB to the top and the other to the bottom branch of this circuit. The top branch multiplies the signal by cos ωt

10

INTRODUCTION

TABLE 1.5 φ

QPSK Scheme

SI

SQ

π/4 1 3π/4 −1 5π/4 −1 7π/4 1

1 1 −1 −1

Dibit Representation 11 01 00 10

Logic 1 = 1 V Logic 0 = −1 V

A cos ωt

Demultiplexer SBB(t)

I

Q

LO

Σ

Smod(t)

90° Phase shift A sin ωt Logic 1 = 1 V Logic 0 = −1 V

Figure 1.8

Block diagram of a QPSK modulation scheme.

Threshold detector

SBB(t)

Integrator A cos ωt

Multiplexer I

Smod(t)

LO

Q

90° Phase shift Asin ωt Threshold detector

Integrator

Figure 1.9 Block diagram of a QPSK demodulation scheme.

and the bottom signal by sin ωt. The outputs of the two mixers are then added to generate Smod . Demodulation inverts the modulation to retrieve SBB . As illustrated in Figure 1.9, Smod is multiplied by cos ωt in the top branch and by sin ωt in the bottom branch of the demodulator. The top integrator stops SQ , and the bottom integrator, SI . Two threshold detectors generate the corresponding logic states that are multiplexed by the multiplexer unit to recover SBB . Chapter 2 provides an overview of wireless communication systems and their characteristics.

2 COMMUNICATION SYSTEMS

Modern communication systems require RF and microwave signals for the wireless transmission of information. These systems employ oscillators, mixers, filters, and amplifiers to generate and process various types of signals. The transmitter communicates with the receiver via antennas placed on each side. Electrical noise associated with the systems and the channel affects the performance. A system designer needs to know about the channel characteristics and system noise in order to estimate the required power levels. The chapter begins with an overview of microwave communication systems and RF wireless services to illustrate the applications of circuits and devices that are described in the following chapters. It also covers the placement of various building blocks in a given system. A short discussion on antennas is included to help in understanding signal behavior when it propagates from a transmitter to a receiver. The Friis transmission formula and the radar range equation are important to understanding the effects of frequency, range, and operating power levels on the performance of a communication system. Note that radar concepts now find many other applications, such as proximity or level sensing in an industrial environment. Therefore, a brief discussion of Doppler radar is also included. Noise and distortion characteristics play a significant role in analysis and design of these systems. Minimum detectable signal (MDS), gain compression, intercept point, and the dynamic range of an amplifier (or receiver) are introduced later. Other concepts associated with noise and distortion characteristics are also introduced in this chapter. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

11

12

COMMUNICATION SYSTEMS

2.1 TERRESTRIAL COMMUNICATION As mentioned in Chapter 1, microwave signals propagate along the line of sight. Therefore, the Earth’s curvature limits the range over which a microwave communication link can be established. A transmitting antenna sitting on a 25-ft-high tower can typically communicate only up to a distance of about 50 km. Repeaters can be placed at regular intervals to extend the range. Figure 2.1 is a block diagram of a typical repeater. A repeater system operates as follows. A microwave signal arriving at antenna A works as input to port 1 of a circulator. It is directed to port 2 without loss, assuming that the circulator is ideal. Then it passes through a receiver protection circuit that limits the magnitude of large signals but passes those of low intensity with negligible attenuation. The purpose of this circuit is to block excessively large signals from reaching the receiver input. The mixer following it works as a down-converter that transforms a high-frequency signal to a low-frequency signal, typically in the range of 70 MHz. A Schottky diode is generally employed in the mixer because of its superior noise characteristics. This frequency conversion facilitates amplification of the signal economically. A bandpass filter is used at the output of the mixer to stop undesired harmonics. An intermediate-frequency (IF) amplifier is then used to amplify the signal. It is generally a low-noise solid-state

Circulator B

Circulator A A

1

Transmitter for direction A

3

Received from direction B

1

2

3

2 Receiver protection circuit

Transreceiver for the reverse direction (from B to A)

Power amplifier

Mixer Mixer

Bandpass filter

IF amplifier with AGC

Limiter circuit

Bandpass filter

Mixer Bandpass filter

Power divider

Shift oscillator

Figure 2.1

Microwave source

Block arrangement of a repeater system.

B

SATELLITE COMMUNICATION

13

amplifier with ultralinear characteristics over a broadband. The amplified signal is mixed with another signal for up-conversion of frequency. After filtering out undesired harmonics introduced by the mixer, it is fed to a power amplifier stage that feeds circulator B for onward transmission through antenna B. This upconverting mixer circuit generally employs a varactor diode. Circulator B directs the signal entering at port 3 to the antenna connected at its port 1. Similarly, the signal propagating upstream is received by antenna B and the circulator directs it toward port 2. It then goes through the processing as described for the downstream signal and is radiated by antenna A for onward transmission. Hence, the downstream signal is received by antenna A and transmitted in the forward direction by antenna B. Similarly, the upstream signal is received by antenna B and forwarded to the next station by antenna A. The two circulators help channel the signal in the correct direction. A parabolic antenna with tapered horn as primary feeder is generally used in microwave links. This type of composite antenna system, known as a hog horn, is fairly common in high-density links because of its broadband characteristics. These microwave links operate in the frequency range 4 to 6 GHz, and signals propagating in two directions are separated by a few hundred megahertz. Since this frequency range overlaps with C-band satellite communication, the interference of these signals needs to be taken into design consideration. A single frequency can be used twice for transmission of information using vertical and horizontal polarization.

2.2 SATELLITE COMMUNICATION The ionosphere does not reflect microwaves as it does RF signals. However, one can place a conducting object (satellite) up in the sky that reflects them back to Earth. A satellite can even improve the signal quality using on-board electronics before transmitting it back. The gravitational force needs to be balanced somehow if this object is to stay in position. An orbital motion provides this balancing force. If a satellite is placed at low altitude, greater orbital force will be needed to keep it in position. These low- and medium-altitude satellites are visible from a ground station only for short periods. On the other hand, satellites placed at an altitude of about 36,000 km over the equator, called geosynchronous or geostationary satellites are visible from their shadows at all times. C-band geosynchronous satellites use between 5725 and 7075 MHz for their uplinks. The corresponding downlinks are between 3400 and 5250 MHz. Table 2.1 lists the downlink center frequencies of a 24-channel transponder. Each channel has a total bandwidth of 40 MHz; 36 MHz of that carries the information, and the remaining 4 MHz is used as a guard band. It is accomplished with a 500-MHz bandwidth using different polarization for the overlapping frequencies. The uplink frequency plan may be found easily after adding 2225 MHz to these downlink frequencies. Figure 2.2 illustrates the simplified block diagram of a C-band satellite transponder. A 6-GHz signal received from the Earth station

14

COMMUNICATION SYSTEMS

TABLE 2.1

C-Band Downlink Transponder Frequencies

Horizontal Polarization Channel

Vertical Polarization

Center Frequency (MHz)

Channel

Center Frequency (MHz)

3720 3760 3800 3840 3880 3920 3960 4000 4040 4080 4120 4160

2 4 6 8 10 12 14 16 18 20 22 24

3740 3780 3820 3860 3900 3940 3980 4020 4060 4100 4140 4180

1 3 5 7 9 11 13 15 17 19 21 23

Mixer BPF

Uplink 6-GHz signal

LNA

BPF

TWT amp.

LO

Figure 2.2

Downlink 4-GHz signal

Simplified block diagram of a transponder.

is passed through a bandpass filter before amplifying it through a low-noise amplifier (LNA). It is then mixed with a local oscillator (LO) signal to bring down its frequency. A bandpass filter that is connected right after the mixer filters out the unwanted frequency components. This signal is then amplified by a traveling wave tube (TWT) amplifier and transmitted back to Earth. Another frequency band in which satellite communication has been growing continuously is the Ku-band. The geosynchronous Fixed Satellite Service (FSS) generally operates between 10.7 and 12.75 GHz (space to Earth) and 13.75 to 14.5 GHz (Earth to space). It offers the following advantages over the C-band: ž ž

The size of the antenna can be smaller (3 ft or even smaller, with higherpower satellites against 8 to 10 ft for C-band). Because of higher frequencies used in the up- and downlinks, there is no interference with C-band terrestrial systems.

Since higher-frequency signals attenuate faster while propagating through adverse weather (rain, fog, etc.), Ku-band satellites suffer from this major

SATELLITE COMMUNICATION

15

drawback. Signals with higher powers may be used to compensate for this loss. Generally, this power is on the order of 40 to 60 W. The high-power direct broadcast satellite (DBS) system uses power amplifiers in the range 100 to 120 W. The National Broadcasting Company (NBC) has been using the Ku-band to distribute the programming to its affiliates. Also, various news-gathering agencies have used this frequency band for some time. Convenience stores, auto parts distributors, banks, and other businesses have used the very small aperture terminal (VSAT) because of its small antenna size (typically, on the order of 3 ft in diameter). It offers two-way satellite communication; usually back to hub or headquarters. The Public Broadcasting Service (PBS) uses VSATs for exchanging information among public schools. Direct broadcast satellites (DBSs) have been around since 1980, but early DBS ventures failed for various reasons. In 1991, Hughes Communications entered into the direct-to-home (DTH) television business. DirecTV was formed as a unit of GM Hughes, with DBS-1 launched in December 1993. Its longitudinal orbit is at 101.2◦ W, and it employs a left-handed circular polarization. DBS-2, launched in August 1994 uses a right-handed circular polarization, and its orbital longitude is at 100.8◦ W. DirecTV employs a digital architecture that can utilize video and audio compression techniques. It complies with Motion Picture Experts Group (MPEG)-2. By using compression ratios of 5 to 7, over 150 channels of programs are available from the two satellites. These satellites include 120-W TWT amplifiers that can be combined to form eight pairs at 240 W of power. This higher power can also be utilized for high-definition television (HDTV) transmission. Earth-to-satellite link frequency is 17.3 to 17.8 GHz; satellite-toEarth link frequency uses the 12.2- to 12.7-GHz band. Circular polarization is used because it is less affected by rain than is linear orthogonal polarization. Several communication services are now available that use low-Earth-orbit satellites (LEOSs) and medium-Earth-orbit satellites (MEOSs). LEOS altitudes range from 750 to 1500 km; MEOS systems have an altitude around 10,350 km. These services compete with or supplement cellular systems and geosynchronous Earth-orbit satellites (GEOSs). The GEOS systems have some drawbacks, due to the large distances involved. They require relatively large powers, and the propagation time delay creates problems in voice and data transmissions. The LEOS and MEOS systems orbit Earth faster because of being at lower altitudes, and these are therefore visible only for short periods. As Table 2.2 indicates, several satellites are used in a personal communication system to solve this problem. Three classes of service can be identified for mobile satellite services: 1. Data transmission and messaging from very small, inexpensive satellites 2. Voice and data communications from big LEOSs 3. Wideband data transmission Another application of L-band microwave frequencies (1227.60 and 1575.42 MHz) is in global positioning systems (GPSs). A constellation of 24 satellites is used to determine a user’s geographical location. Two services

16

COMMUNICATION SYSTEMS

TABLE 2.2

Specifications of Certain Personal Communication Satellites Iridium (LEO)

Number of satellites Altitude (km) Uplink (GHz) Downlink (GHz) Gateway terminal uplink (GHz) Gateway terminal downlink (GHz) Average satellite connect time (min) Features of handset Modulation BER Supportable data rate (kb/s)

Globalstar (LEO)

Odyssey (MEO)

66 755 1.616–1.6265 1.616–1.6265 27.5–30.0

48 1,390 1.610–1.6265 2.4835–2.500 C-band

12 10,370 1.610–1.6265 2.4835–2.500 29.5–30.0

18.8–20.2

C-band

19.7–20.2

9

10–12

120

QPSK 1E-2 (voice) 1E-5 (data) 4.8 (voice) 2.4 (data)

FQPSK 1E-3 (voice) 1E-5 (data) 1.2–9.6 (voice and data)

QPSK 1E-3 (voice) 1E-5 (data) 4.8 (voice) 1.2–9.6 (data)

are available: the standard positioning service (SPS) for civilian use, utilizing a single-frequency course/acquisition (C/A) code, and the precise positioning service (PPS) for the military, utilizing a dual-frequency P-code (protected). These satellites are at an altitude of 10,900 miles above the Earth, with an orbital period of 12 hours.

2.3 RADIO-FREQUENCY WIRELESS SERVICES A lot of exciting wireless applications have been reported that use voice and data communication technologies. Wireless communication networks consist of microcells that connect people with truly global, pocket-size communication devices, telephones, pagers, personal digital assistants, and modems. Typically, a cellular system employs a 100-W transmitter to cover a cell 0.5 to 10 miles in radius. The handheld transmitter has a power of less than 3 W. Personal communication networks (PCN/PCS) operate with a 0.01- to 1-W transmitter to cover a cell radius of less than 450 yards. The handheld transmitter power is typically less than 10 mW. Table 2.3 shows the cellular telephone standards of selected systems. There have been no universal standards set for wireless personal communication. In North America, cordless has been CT-0 (an analog 46/49-MHz standard) and cellular AMPS (Advanced Mobile Phone Service) operating at 800 MHz. The situation in Europe has been far more complex; every country has had its own standard. Although cordless was nominally CT-0, different countries used their own frequency plans. This led to a plethora of new standards. These include,

17

RADIO-FREQUENCY WIRELESS SERVICES

TABLE 2.3

Selected Cellular Telephones Analog Cellular Phone Standard NADC (IS-54)

NADC (IS-95)

GSM

PDC

824–849 871–904

824–849

824–849

880–915

869–894 916–949

869–894

869–894

925–960

940–956 1477–1501 810–826 1429–1453

600 mW

200 mW

1W

AMP Freq. range (MHz) Tx Rx Transmitter power (max.) Multiple access Number of channels Channel spacing (kHz) Modulation Bit rate (kb/s)

TABLE 2.4

Digital Cellular Phone Standard

ETACS

FDMA 832 30

FDMA TDMA/FDM CDMA/FDM TDMA/FDM TDMA/FDM 1000 832 20 124 1600 25 30 1250 200 25

FM —

FM —

π/4 DQPSK QPSK/BPSK 48.6 1228.8

π/4 DQPSK 42

GMSK 270.833

Selected Cordless Telephones Analog Cordless Phone Standard CT-0

CT-1 and CT-1+

Frequency range (MHz)

46/49

CT-1: 914/960; CT-1+: 885–932

Transmitter power (max.) (mW) Multiple access Number of channels Channel spacing (kHz) Modulation Bit rate (kb/s)

—

—

FDMA FDMA 10–20 CT-1: 40; CT-1+: 80 40 25 FM

FM

—

—

Digital Cordless Phone Standard CT-2 and CT-2+ CT-1: 864–868; CT-2+: 930/931; 940/941 10 and 80

TDMA/FDM 40

DECT

PHS (Formerly PHP)

1880–1900

1895–1918

250

80

100

TDMA/FDM TDMA/FDM 10 (12 users 300 (four users per channel) per channels) 1728 300

GFSK 72

GFSK 1152

π/4 DQPSK 384

but are not limited to, CT-1, CT-1+, DECT (Digital European Cordless Telephone), PHP (Personal Handy Phone in Japan), E-TACS (Extended Total Access Communication System in the UK), NADC (North American Digital Cellular), GSM (Global System for Mobile Communication), and PDC (Personal Digital Cellular). Specifications for selected cordless telephones are given in Table 2.4.

18

COMMUNICATION SYSTEMS

2.4 ANTENNA SYSTEMS Figure 2.3 illustrates some of the antennas that are used in communication systems. These can be categorized into two groups: wire antennas and aperture-type antennas. Electric dipole, monopole, and the loop antennas belong to the former group; horn, reflector, and lens belong to the latter category. The aperture antennas can be further subdivided into primary and secondary (or passive) antennas. Primary antennas are directly excited by the source and can be used independently for transmission or reception of signals. On the other hand, a secondary antenna requires another antenna as its feeder. Horn antennas fall in the first category, whereas the reflector and lens belong to the second. Various types of horn antennas are commonly used as feeders in reflector and lens antennas.

Electronics

Electronics (a)

(b)

(c)

Transmitter

(d )

(e)

Transmitter

(f)

Figure 2.3 Some commonly used antennas: (a) electric dipole; (b) monopole; (c) loop; (d) pyramidal horn; (e) Cassegrain reflector; (f ) lens.

ANTENNA SYSTEMS

19

When an antenna is energized, it generates two types of electromagnetic fields. Part of the energy stays nearby and part propagates outward. The propagating signal represents the radiation fields, while the nonpropagating is reactive (capacitive or inductive) in nature. Space surrounding the antenna can be divided into three regions. The reactive fields dominate in the nearby region but are reduced in strength at a faster rate than those associated with the propagating signal. If the largest dimension of an antenna is D and the signal wavelength is λ, reactive fields dominate up to about 0.62 D 3 /λ and diminish after 2D 2 /λ. The region beyond 2D 2 /λ is called the far-field (or radiation field) region. Power radiated by an antenna per unit solid angle is known as the radiation intensity U . It is a far-field parameter that is related to power density (power per unit area) Wrad and distance r as follows: U = r 2 Wrad

(2.4.1)

Directive Gain and Directivity If an antenna radiates uniformly in all directions, it is called an isotropic antenna. This is a hypothetical antenna that helps in defining the characteristics of a real one. The directive gain DG is defined as the ratio of radiation intensity due to the test antenna to that of an isotropic antenna. It is assumed that total radiated power remains the same in the two cases. Hence, DG =

U 4πU = U0 Prad

(2.4.2)

where U is the radiation intensity due to the test antenna in watts per unit solid angle, U0 the radiation intensity due to the isotropic antenna in watts per unit solid angle, and Prad the total power radiated in watts. Since U is a directionaldependent quantity, the directive gain of an antenna depends on the angles θ and φ. If the radiation intensity assumes its maximum value, the directive gain is called the directivity D0 . That is, D0 =

Umax 4πUmax = U0 Prad

(2.4.3)

Gain of an Antenna The power gain of an antenna is defined as the ratio of its radiation intensity at a point to the radiation intensity that results from a uniform radiation of the same input power. Hence, gain = 4π

radiation intensity U (θ, φ) = 4π total input power Pin

(2.4.4)

Most of the time we deal with relative gain. It is defined as a ratio of the power gain of the test antenna in a given direction to the power gain of a reference

20

COMMUNICATION SYSTEMS

antenna. Both antennas must have the same input power. The reference antenna is usually a dipole, horn, or any other antenna whose gain can be calculated or is known. However, the reference antenna is a lossless isotropic radiator in most cases. Hence, U (θ, φ) gain = 4π (2.4.5) Pin (lossless isotropic antenna) When the direction is not stated, the power gain is usually taken in the direction of maximum radiation. Radiation Patterns and Half-Power Beam Width Far-field power distribution at a distance r from the antenna depends on the spatial coordinates θ and φ. Graphical representations of these distributions on the orthogonal plane (θ-plane or φ-plane) at a constant distance r from the antenna are called its radiation patterns. Figure 2.4 illustrates the radiation pattern of the vertical dipole antenna with θ. Its φ-plane pattern can be found after rotating it about the vertical axis. Thus, a three-dimensional picture of the radiation pattern of a dipole is doughnut shaped. Similarly, the power distributions of other antennas generally show peaks and valleys in the radiation zone. The highest peak between the two valleys is known as the main lobe; the others are called side lobes. The total angle about the main peak over which power is reduced by 50% of its maximum value is called the half-power beam width on that plane.

θ

r 0

1

Figure 2.4 Radiation pattern of a dipole in a vertical (θ) plane.

ANTENNA SYSTEMS

21

The following relations are used to estimate the power gain G and the halfpower beam width (HPBW, or simply BW) of an aperture antenna: G=

4π 4π Ae = 2 Aκ 2 λ λ

(2.4.6)

65λ d

(2.4.7)

and BW(degrees) =

where Ae is the effective area of radiating aperture in square meters, A its physical area (πd 2 /4 for a reflector antenna dish with diameter d), κ the antenna efficiency (ranges from 0.6 to 0.65), and λ the signal wavelength in meters. Example 2.1 Calculate the power gain (in decibels) and the half-power beam width of a parabolic dish antenna 30 m in diameter that is radiating at 4 GHz. SOLUTION The signal wavelength and area of the aperture are λ=

3 × 108 = 0.075 m 4 × 109

A=

πd 2 302 =π = 706.8584 m2 4 4

and

Assuming that the aperture efficiency is 0.6, the antenna gain and half-power beam width are found as follows: G=

4π × 706.8584 × 0.6 = 947,482.09 (0.075)2 = 10 log10 (947,482.09) = 59.76 ≈ 60 dB

BW =

65 × 0.075 ◦ = 0.1625 30

Antenna Efficiency If an antenna is not matched with its feeder, a part of the signal available from the source is reflected back. It is considered to be the reflection (or mismatch) loss. The reflection (or mismatch) efficiency is defined as a ratio of power input to the antenna to that of power available from the source. Since the ratio of reflected power to that of power available from the source is equal to the square of the magnitude of the voltage reflection coefficient, the reflection efficiency er is given by er = 1 − ||2

22

COMMUNICATION SYSTEMS

where = voltage reflection coefficient =

ZA − Z0 ZA + Z0

Here ZA is the antenna impedance and Z0 is the characteristic impedance of the feeding line. In addition to mismatch, the signal energy may dissipate in an antenna due to imperfect conductor or dielectric material. These efficiencies are hard to compute. However, the combined conductor and dielectric efficiency ecd can be determined experimentally after measuring the input power Pin and the radiated power Prad . It is given as Prad ecd = Pin The overall efficiency eo is a product of the efficiencies above. That is, eo = er ecd

(2.4.8)

Example 2.2 A 50- transmission line feeds a lossless one-half-wavelengthlong dipole antenna. The antenna impedance is 73 . If its radiation intensity, U (θ, φ), is given as U = B0 sin3 θ find the maximum overall gain. SOLUTION The maximum radiation intensity, Umax , is the B0 value that occurs at θ = π/2. Its total radiated power is found as follows:

2π

Prad = 0

π 0

B0 sin3 θ sin θ dθ dφ = 34 π2 B0

Hence, D0 = 4π

Umax 4πB0 16 = 1.6977 = 3 2 = Prad 3π π B0 4

or D0 (dB) = 10 log10 (1.6977)dB = 2.2985 dB Since the antenna is lossless, the radiation efficiency ecd is unity (0 dB). Its mismatch efficiency is computed as follows. The voltage reflection coefficient at its input (it is formulated in Chapter 2) is =

23 ZA − Z0 73 − 50 = = ZA + Z0 73 + 50 123

ANTENNA SYSTEMS

23

Therefore, the mismatch efficiency of the antenna is er = 1 − (23/123)2 = 0.9650 = 10 log10 (0.9650)dB = −0.1546 dB The overall gain G0 (in decibels) is found as follows: G0 (dB) = 2.2985 − 0 − 0.1546 = 2.1439 dB Bandwidth Antenna characteristics such as gain, radiation pattern, impedance, and so on are frequency dependent. The bandwidth of an antenna is defined as the frequency band over which its performance with respect to some characteristic (HPBW, directivity, etc.) conforms to a specified standard. Polarization Polarization of an antenna is the same as polarization of its radiating wave. It is a property of the electromagnetic wave describing the time-varying direction and relative magnitude of an electric field vector. The curve traced by the instantaneous electric field vector with time is the polarization of that wave. The polarization is classified as follows: ž ž

ž

Linear polarization. If the tip of the electric field intensity traces a straight line in some direction with time, the wave is linearly polarized. Circular polarization. If the end of the electric field traces a circle in space as time passes, that electromagnetic wave is circularly polarized. Further, it may be right-handed circularly polarized (RHCP) or left-handed circularly polarized (LHCP), depending on whether the electric field vector rotates clockwise or counterclockwise. Elliptical polarization. If the tip of the electric field intensity traces an ellipse in space as time lapses, the wave is elliptically polarized. As in the preceding case, it may be right- or left-handed elliptical polarization (RHEP and LHEP).

In a receiving system, the polarization of the antenna and the incoming wave need to be matched for maximum response. If this is not the case, there will be some signal loss, known as polarization loss. For example, if there is a vertically polarized wave incident on a horizontally polarized antenna, the induced voltage available across its terminals will be zero. In this case, the antenna is crosspolarized with an incident wave. The square of the cosine of the angle between wave polarization and antenna polarization is a measure of the polarization loss. It can be determined by squaring the scalar product of unit vectors representing the two polarizations.

24

COMMUNICATION SYSTEMS

Example 2.3 The electric field intensity of an electromagnetic wave propagating in a lossless medium in the z-direction is given by E(r, t) = xE ˆ 0 (x, y) cos(ωt − kz)

V/m

It is incident upon an antenna that is linearly polarized as follows: Ea (r) = (xˆ + y)E(x, ˆ y, z)

V/m

Find the polarization loss factor. SOLUTION In this case, the incident wave is linearly polarized along the x-axis while the receiving antenna is linearly polarized at 45◦ from it. Therefore, onehalf of the incident signal is cross-polarized with the antenna. It is determined mathematically as follows. The unit vector along the polarization of incident wave is uˆ i = xˆ The unit vector along the antenna polarization may be found as 1 uˆ a = √ (xˆ + y) ˆ 2 Hence, the polarization loss factor is |uˆ i •uˆ a |2 = 0.5 = −3.01 dB Effective Isotropic Radiated Power The effective isotropic radiated power (EIRP) is a measure of the power gain of the antenna. It is equal to the power needed by an isotropic antenna that provides the same radiation intensity at a given point as that of the directional antenna. If power input to the feeding line is Pt and the antenna gain is Gt , the EIRP is defined as Pt Gt EIRP = (2.4.9) L where L is the input/output power ratio of the transmission line that is connected between the output of the final power amplifier stage of the transmitter and the antenna. It is given by Pt L= (2.4.10) Pant Alternatively, the EIRP can be expressed in dBw as EIRP(dBw) = Pt (dBw) − L(dB) + G(dB)

(2.4.11)

ANTENNA SYSTEMS

25

Example 2.4 In a transmitting system, the output of the final high-power amplifier is 500 W, and the line feeding its antenna has an attenuation of 20%. If the gain of the transmitting antenna is 60 dB, find the EIRP in dBw. SOLUTION Pt = 500 W = 26.9897 dBw Pant = 0.8 × 500 = 400 W G = 60 dB = 106 and L=

500 = 1.25 = 10 log10 (1.25) = 0.9691 dB 400

Hence, EIRP(dBw) = 26.9897 − 0.9691 + 60 = 86.0206 dBw or EIRP =

500 × 106 = 400 × 106 W 1.25

Space Loss The transmitting antenna radiates in all directions, depending on its radiation characteristics. However, the receiving antenna receives only the power that is incident on it. Hence, the rest of the power is not used and is lost in space. It is represented by the space loss. It can be determined as follows. The power density wt of a signal transmitted by an isotropic antenna is given by Pt wt = W/m2 (2.4.12) 4πR 2 where Pt is the transmitted power in watts and R is the distance from the antenna in meters. The power received by a unity-gain antenna located at R is found to be Pr = wt Aeu where Aeu is the effective area of an isotropic antenna. From (2.4.6), for an isotropic antenna, G=

4π Aeu = 1 λ2

or Aeu =

λ2 4π

(2.4.13)

26

COMMUNICATION SYSTEMS

Hence, (2.4.12) can be written as Pr =

Pt λ2 4πR 2 4π

(2.4.14)

and the space loss ratio is found to be Pr = Pt

λ 4πR

2 (2.4.15)

It is usually expressed in decibels as follows: space loss ratio = 20 log10

λ 4πR

dB

(2.4.16)

Example 2.5 A geostationary satellite is 35,860 km away from Earth’s surface. Find the space loss ratio if it is operating at 4 GHz. SOLUTION R = 35,860,000 m

and λ =

3 × 108 = 0.075 m 4 × 109

Hence, space loss ratio =

4π × 35,860,000 0.075

2

= 2.77 × 10−20 = −195.5752 dB

Friis Transmission Formula and Radar Range Equation Analysis and design of communication and monitoring systems often require an estimation of transmitted and received powers. The Friis transmission formula and radar range equation provide the means for such calculations. The former is applicable to a one-way communication system where the signal is transmitted at one end and is received at the other end of the link. The radar range equation is applicable when the signal transmitted hits a target and the signal reflected is generally received at the location of the transmitter. We consider these two formulations next. Friis Transmission Equation Consider a simplified communication link as illustrated in Figure 2.5. A distance R separates the transmitter and the receiver. Effective apertures of transmitting and receiving antennas are Aet and Aer , respectively. Further, the two antennas

27

ANTENNA SYSTEMS

R

Transmitter

Figure 2.5

Receiver

Simplified block diagram of a communication link.

are assumed to be polarization matched. If power input to the transmitting antenna is Pt , isotropic power density w0 at a distance R from the antenna is given as w0 =

Pt et 4πR 2

(2.4.17)

where et is the radiation efficiency of the transmitting antenna. For a directional transmitting antenna, the power density wt can be written as wt =

Pt Gt Pt et Dt = 4πR 2 4πR 2

(2.4.18)

where Gt is the gain and Dt is the directivity of transmitting antenna. The power collected by the receiving antenna is Pr = Aer wt

(2.4.19)

λ2 Gr 4π

(2.4.20)

From (2.4.6), Aer =

where the receiving antenna gain is Gr . Therefore, we find that Pr = or Pr = Pt

λ2 λ2 Pt Gt Gr wt = Gr 4π 4π 4πR 2

λ 4πR

2 Gr Gt = er et

λ 4πR

2 Dr Dt

(2.4.21)

If signal frequency is f , for a free-space link, λ 3 × 108 = 4πR 4πf R where f is in hertz and R is in meters.

(2.4.22)

28

COMMUNICATION SYSTEMS

Generally, the link distance is long and the signal frequency is high, such that kilometer and megahertz will be more convenient units than the usual meter and hertz, respectively. For R in kilometers and f in megahertz, we find that λ 0.3 3 × 108 1 = = 6 3 4πR 4π × 10 fMHz × 10 Rkm 4π fMHz Rkm

(2.4.23)

Hence, from (2.4.21), Pr (dBm) = Pt (dBm) + 20 log10

0.3 − 20 log10 (fMHz Rkm ) + Gt (dB) + Gr (dB) 4π

or Pr (dBm) = Pt (dBm) + Gt (dB) + Gr (dB) − 20 log10 (fMHz Rkm ) − 32.4418 (2.4.24) where the power transmitted and the power received are in dBm while the two antenna gains are in decibels. Example 2.6 A 20-GHz transmitter on board the satellite uses a parabolic antenna that is 45.7 cm in diameter. The antenna gain is 37 dB and its radiated power is 2 W. The ground station that is 36,941.031 km away from it has an antenna gain of 45.8 dB. Find the power collected by the ground station. How much power would be collected at the ground station if there were isotropic antennas on both sides? SOLUTION The power transmitted, Pt (dBm) = 10 log10 (2000) = 33.0103 dBm and 20 log10 (fMHz Rkm ) = 20 log10 (20 × 103 × 36,941.031) = 177.3708 dB Hence, the power received at the Earth station is found as follows: Pr (dBm) = 33.0103 + 37 + 45.8 − 177.3708 − 32.4418 = −94.0023 dBm or Pr = 3.979 × 10−10 mW If the two antennas are isotropic, Gt = Gr = 1 (or 0 dB) and therefore Pr (dBm) = 33.0103 + 0 + 0 − 177.3708 − 32.4418 = −176.8023 dBm or Pr = 2.0882 × 10−18 mW

29

ANTENNA SYSTEMS

Radar Equation In the case of a radar system, the signal transmitted is scattered by the target in all possible directions. The receiving antenna collects part of the energy that is scattered back toward it. Generally, a single antenna is employed for both the transmitter and the receiver, as shown in Fig. 2.6. If power input to the transmitting antenna is Pt and its gain is Gt , the power density winc incident on the target is Pt Gt Pt Aet winc = = 2 2 (2.4.25) 2 4πR λ R where Aet is the effective aperture of the transmitting antenna. The radar cross section σ of an object is defined as the area intercepting that amount of power that when scattered isotropically produces at the receiver a power density that is equal to that scattered by the actual target. Hence, radar cross section =

scattered power incident power density

or σ=

4πr 2 wr winc

square meters

(2.4.26)

where wr is isotropically backscattered power density at a distance r and winc is power density incident on the object. Hence, the radar cross section of an object is its effective area that intercepts an incident power density winc and gives an isotropically scattered power of 4πr 2 wr for a backscattered power density. Radar cross sections of selected objects are listed in Table 2.5. Using the radar cross section of a target, the power intercepted by it can be found as follows: σPt Gt Pinc = σwinc = (2.4.27) 4πR 2 Power density arriving back at the receiver is wscatter =

Pinc 4πR 2

Transmitter

Circulator Receiver

Figure 2.6

Antenna

Radar system.

(2.4.28)

30

COMMUNICATION SYSTEMS

TABLE 2.5

Radar Cross Sections of Selected Objects Radar Cross Section (m2 )

Object Pickup truck Automobile Jumbo-jet airliner Large bomber Large fighter aircraft Small fighter aircraft Adult male Conventional winged missile Bird Insect Advanced tactical fighter

200 100 100 40 6 2 1 0.5 0.01 0.00001 0.000001

and power available at the receiver input is Pr = Aer wscatter =

Gr λ2 σPt Gt σAer Aet Pt = 2 2 4π(4πR ) 4πλ2 R 4

(2.4.29)

Example 2.7 A distance of 100λ separates two lossless X-band horn antennas (Figure 2.7). Reflection coefficients at the terminals of transmitting and receiving antennas are 0.1 and 0.2, respectively. Maximum directivities of the transmitting and receiving antennas are 16 and 20 dB, respectively. Assuming that the input power in a lossless transmission line connected to the transmitting antenna is 2 W and that the two antennas are aligned for maximum radiation between them and are polarization matched, find the power input to the receiver. SOLUTION As discussed in Chapter 3, impedance discontinuity generates an echo signal very similar to that of an acoustical echo. Hence, signal power available beyond the discontinuity is reduced. The ratio of the reflected signal voltage to that of the incident is called the reflection coefficient. Since the power is proportional to the square of the voltage, the power reflected from the discontinuity is equal to the square of the reflection coefficient times the incident power.

100l Receiver

Transmitter

Pd = [1 − |Γ|2]PR Pt = [1 − |Γ|2]PS

Figure 2.7

Setup for Example 2.7.

ANTENNA SYSTEMS

31

Therefore, power transmitted in the forward direction will be given by Pt = (1 − ||2 )Pin and the power radiated by the transmitting antenna is found to be Pt = (1 − 0.12 )2 = 1.98 W Since the Friis transmission equation requires the antenna gain as a ratio instead of in decibels, Gt and Gr are calculated as follows: Gt = 16 dB = 101.6 = 39.8107

and Gr = 20 dB = 102.0 = 100

Hence, from (2.4.21), Pr =

λ 4π × 100λ

2 × 100 × 39.8107 × 1.98

or Pr = 5 mW and power delivered to the receiver, Pd , is Pd = (1 − 0.22 )5 = 4.8 mW Example 2.8 Radar operating at 12 GHz transmits 25 kW through an antenna of 25 dB gain. A target with its radar cross section at 8 m2 is located at 10 km from the radar. If the same antenna is used for the receiver, determine the power received. SOLUTION Pt = 25 kW f = 12 GHz → λ =

3 × 108 = 0.025 m 12 × 109

Gr = Gt = 25 dB → 102.5 = 316.2278 R = 10 km

σ = 8 m2

Hence, Pr =

Gr Gt Pt σλ2 316.22782 × 25, 000 × 8 × 0.0252 = = 6.3 × 10−13 W 4π(4πR 2 )2 (4π)3 (104 )4

or Pr = 0.63 pW

32

COMMUNICATION SYSTEMS

Doppler Radar An electrical signal propagating in free space can be represented by a simple expression as follows: v(z, t) = A cos(ωt − kz) (2.4.30) The signal frequency is ω radians per second and k is its wave number (equal to ω/c, where c is the speed of light in free space) in radians per meter. Assume that there is a receiver located at z = R, as shown in Figure 2.5 and R is changing with time (the receiver may be moving toward or away from the transmitter). In this situation, the receiver response vo (t) is given as follows: vo (t) = V cos(ωt − kR)

(2.4.31)

The angular frequency, ω0 , of vo (t) can be determined easily after differentiating the argument of the cosine function with respect to time. Hence, ω0 =

d dR (ωt − kR) = ω − k dt dt

(2.4.32)

Note that k is time independent and that the time derivative of R represents the velocity, vr , of the receiver with respect to the transmitter. Hence, (2.4.32) can be written ωvr vr (2.4.33) ω0 = ω − =ω 1− c c If the receiver is closing in, vr will be negative (negative slope of R), and therefore the signal received will indicate a signal frequency higher than ω. On the other hand, it will show a lower frequency if R is increasing with time. It is the Doppler frequency shift that is employed to design the Doppler radar. Consider the simplified block diagram shown in Figure 2.8. A microwave signal generated by the oscillator is split into two parts via a power divider. The circulator feeds one part of this power to an antenna that illuminates a target while a mixer uses the remaining fraction as its reference signal. Further, the antenna intercepts a part of the signal that is scattered by an object. It is then directed to the mixer through a circulator. The mixer output includes a difference frequency signal that can be filtered out for further processing. Two inputs to the mixer will have the same frequency if the target is stationary, and therefore the Doppler shift δω will be zero. On the other hand, the mixer output will have Doppler frequency if the target is moving. Note that the signal travels twice over the same distance, and therefore the Doppler frequency shift in this case will be twice that found via (2.4.33). Mathematically,

2vr ω0 = ω 1 − c

(2.4.34)

33

NOISE AND DISTORTION Circulator Oscillator

ω

ω

Power divider

ω0

ω ω0

Mixer δω

Amplifier

Display

Figure 2.8 Simplified block diagram of a Doppler radar.

and δω =

2ωvr c

(2.4.35)

2.5 NOISE AND DISTORTION Random movement of charges or charge carriers in an electronic device generates currents and voltages that vary randomly with time. In other words, the amplitude of these electrical signals cannot be predicted at any time. However, it can be expressed in terms of probability density functions. These signals are termed noise. For most applications it suffices to know the mean-square or root-meansquare value. Since the square of the current or the voltage is proportional to the power, mean-square noise voltage and current values are generally called noise power. Further, noise power is normally a function of frequency and the power per unit frequency (watts per hertz) is defined as the power spectral density of noise. If the noise power is the same over the entire frequency band of interest, it is called white noise. There are several mechanisms that can cause noise in an electronic device. Some of these are as follows: ž

ž

Thermal noise. This is the most basic type of noise, which is caused by thermal vibration of bound charges. Johnson studied this phenomenon in 1928, and Nyquist formulated an expression for spectral density around the same time. Therefore, it is also known as Johnson noise or Nyquist noise. In most electronic circuits, thermal noise dominates; therefore, it will be described further because of its importance. Shot noise. This is due to random fluctuations of charge carriers that pass through the potential barrier in an electronic device. For example, electrons

34

COMMUNICATION SYSTEMS

emitted from the cathode of thermionic devices or charge carriers in Schottky diodes produce a current that fluctuates about the average value I . The mean-square current due to shot noise is generally given by 2 iSh = 2eIB

ž

(2.5.1)

where e is electronic charge (1.602 × 10−19 C) and B is the bandwidth in hertz. Flicker noise. This occurs in solid-state devices and vacuum tubes operating at low frequencies. Its magnitude decreases with an increase in frequency. It is generally attributed to chaos in the dynamics of a system. Since the flicker noise power varies inversely with frequency, it is often called 1/f noise. Sometimes it is referred to as pink noise.

Thermal Noise Consider a resistor R that is at a temperature of T Kelvin. Electrons in this resistor are in random motion with a kinetic energy that is proportional to the temperature T . These random motions produce small, random voltage fluctuations across its terminals. This voltage has a zero average value but a nonzero mean-square value vn2 . It is given by Planck’s blackbody radiation law as vn2 =

4hfRB exp(hf/kT ) − 1

(2.5.2)

where h is Planck’s constant (6.546 × 10−34 J · s), k the Boltzmann constant (1.38 × 10−23 J/K), T the temperature in kelvin, B the system bandwidth in hertz, and f the center frequency of the bandwidth in hertz. For frequencies below 100 GHz, the product hf will be smaller than 6.546 × 10−23 J and kT will be greater than 1.38 × 10−22 J if T stays above 10 K. Therefore, kT will be larger than hf for such cases. Hence, the exponential term in equation (2.5.2) can be approximated as follows:

hf exp kT

≈1+

hf kT

Therefore, vn2 ≈

4hf RB = 4BRkT hf/kT

(2.5.3)

This is known as the Rayleigh–Jeans approximation. A Th´evenin-equivalent circuit can replace the noisy resistor as shown in Figure 2.9. As illustrated, it consists of a noise-equivalent voltage source in series with a noise-free resistor. This source will supply a maximum power to a load

35

NOISE AND DISTORTION

R Noise-free resistor

R Noisy resistor

Figure 2.9 Noise-equivalent circuit of a resistor.

of resistance R. The power delivered to that load in a bandwidth B is found as follows: v 2 Pn = n = kT B (2.5.4) 4R Conversely, if an arbitrary white noise source with its driving point impedance R delivers a noise power Ps to a load R, it can be represented by a noisy resistor of value R that is at temperature Te . Hence, Te =

Ps kB

(2.5.5)

where Te is an equivalent temperature selected so that the same noise power is delivered to the load. Consider the noisy amplifier as shown in Figure 2.10. Its gain is G over the bandwidth B. Let the amplifier be matched to the noiseless source and the load resistors. If the source resistor is at a hypothetical temperature of Ts = 0 K, the power input to the amplifier Pi will be zero and the output noise power Po will only be due to noise generated by the amplifier. We can obtain the same noise power at the output of an ideal noiseless amplifier by raising the temperature Ts of the source resistor to Te : Po Te = (2.5.6) GkB Hence, the output power in both cases is Po = GkTe B. The temperature Te is known as the equivalent noise temperature of the amplifier.

Ts = 0 K

R

Noisy amplifier

R

Te =

Po GkB

R

Noiseless amplifier

Figure 2.10 Noise-equivalent representation of an amplifier.

R

36

COMMUNICATION SYSTEMS

Measurement of Noise Temperature by the Y -Factor Method According to the definition, the noise temperature of an amplifier (or any other two-port network) can be determined by setting the source resistance R at 0 K and then measuring the output noise power. However, a temperature of 0 K cannot be achieved in practice. We can circumvent this problem by repeating the experiment at two different temperatures. This procedure is known as the Y-factor method. Consider an amplifier with a power gain of G over a frequency band of B hertz. Further, its equivalent noise temperature is Te (K). The input port of the amplifier is terminated by a matched resistor R while a matched power meter is connected at its output, as illustrated in Figure 2.11. With R at temperature Th , the power meter measures the noise output as P1 . Similarly, the noise power is found to be P2 when the temperature of R is set at Tc . Hence, P1 = GkTh B + GkTe B and P2 = GkTc B + GkTe B For Th higher than Tc , the noise power P1 will be larger than P2 . Therefore, P1 Th + Te =Y = P2 Tc + Te or Te =

Th − Y Tc Y −1

(2.5.7)

For Th larger than Tc , Y will be greater than unity. Further, measurement accuracy is improved by selecting two temperature settings that are far apart. Therefore, Th and Tc represent hot and cold temperatures, respectively. Example 2.9 An amplifier has a power gain of 10 dB in the 500-MHz to 1.5GHz frequency band. The following data are obtained for this amplifier using the

Th Amplifier P1

R

G, B, Te Tc

P2

R

Figure 2.11 Experimental setup for measurement of noise temperature.

NOISE AND DISTORTION

37

Y -factor method: At Th = 290 K, P1 = −70 dBm; at Tc = 77 K, P2 = −75 dBm. Determine its equivalent noise temperature. If this amplifier is used with a source that has an equivalent noise temperature of 450 K, find the output noise power in dBm. SOLUTION Since P1 and P2 are given in dBm, the difference of these two values will give Y in decibels. Hence, Y = (P1 − P2 ) dBm = (−70) − (−75) = 5 dB or Y = 100.5 = 3.1623 Therefore, Te =

290 − (3.1623)(77) = 21.51 K 3.1623 − 1

If a source with an equivalent noise temperature of Ts = 450 K drives the amplifier, the noise power input to this will be kTs B. The total noise power at the output of the amplifier will be Po = GkTs B + GkTe B = 10 × 1.38 × 10−23 × 109 × (450 + 21.51) = 6.5068 × 10−11 W Therefore,

Po = 10 log(6.5068 × 10−8 ) = −71.8663 dBm

Noise Factor and Noise Figure The noise factor of a two-port network is obtained by dividing the signal-to-noise ratio at its input port by the signal-to-noise ratio at its output. Hence, noise factor, F =

Si /Ni So /No

where Si , Ni , So , and No represent the power in input signal, input noise, output signal, and output noise, respectively. If the two-port network is noise-free, the signal-to-noise ratio at its output will be the same as its input, resulting in a noise factor of unity. In reality, the network will add its own noise, while the input signal and noise will be altered by the same factor (gain or loss). It will lower the output signal/noise ratio, resulting in a higher noise factor. Therefore, the noise factor of a two-port network is generally greater than unity. By definition, the input noise power is assumed to be the noise power resulting from a matched resistor at T0 = 290 K (i.e., Ni = kT0 B). Using the circuit arrangement illustrated

38

COMMUNICATION SYSTEMS T0 = 290 K

R Noisy two-port network G, B, Te Pi = Si + Ni

R Po = So + No

Figure 2.12 Circuit arrangement for determination of the noise factor of a noisy two-port network.

in Figure 2.12, the noise factor of a noisy two-port network can be defined as follows: noise factor =

total output noise in B when input source temperature is 290 K output noise of source (only) at 290 K

or F =

Pint kT0 BG + Pint =1+ kT0 BG kT0 BG

(2.5.8)

where Pint represents the output noise power that is generated by the two-port network internally. It can be expressed in terms of noise factor as follows: Pint = k(F − 1)T0 BG = kTe BG

(2.5.9)

where Te is known as the equivalent noise temperature of a two-port network. It is related to the noise factor as follows: Te = (F − 1)T0

(2.5.10)

When the noise factor is expressed in decibels, it is commonly called the noise figure (NF). Hence, NF = 10 log10 F dB (2.5.11) Example 2.10 The power gain of an amplifier is 20 dB in the frequency band 10 to 12 GHz. If its noise figure is 3.5 dB, find the output noise power in dBm. SOLUTION Output noise power = kT0 BG + Pint = F kT0 BG = No F = 3.5 dB = 100.35 = 2.2387 G = 20 dB = 102 = 100 Therefore, No = 2.2387 × 1.38 × 10−23 × 290 × 2 × 109 × 100 = 1.7919 × 10−9 W

39

NOISE AND DISTORTION

or No = 10 log10 (1.7919 × 10−6 mW) dBm = −57.4669 dBm ≈ −57.5 dBm Noise in Two-Port Networks Consider the noisy two-port network shown in Figure 2.13(a). Ys = Gs + j Bs is the source admittance connected at port 1 of the network. The noise it generates is represented by the current source is with a root-mean-square value of Is . This noisy two-port can be replaced by a noise-free network with a current source in and a voltage source vn connected at its input, as shown in Figure 2.13(b). In and Vn represent the corresponding root-mean-square current and voltage of the noise. It is assumed that the noise represented by is is uncorrelated with that represented by in and vn . However, a part of in , inc , is assumed to be correlated with vn via the correlation admittance Yc = Gc + j Xc while the remaining part inu is uncorrelated. Hence, Is2 = is2 = 4kTBG S Vn2

=

vn2

(2.5.12)

= 4kTBR n

(2.5.13)

and 2 2 = inu = 4kTBG nu Inu

(2.5.14)

I1

↑

Is

Ys

I2

Noisy two-port network

V1

V2

(a) I1

Is

↑

Ys

I2

Vn

V1

In

↑

Noise-free two-port network

(b)

Figure 2.13 Noisy two-port network (a) and its equivalent circuit (b).

V2

40

COMMUNICATION SYSTEMS

Now we find a Norton equivalent for the circuit that is connected at the input of the noise-free two-port network shown in Figure 2.13(b). Since these are random 2 variables, the mean-square noise current < ieq > is found as follows: 2 = is2 + |in + Ys vn |2 = is2 + |inu + inc + Ys vn |2 ieq

= is2 + |inu + (Yc + Ys )vn |2 or 2 2 ieq = is2 + inu + |Yc + Yn |2 vn2

(2.5.15)

Hence, the noise factor F is F =

2 ieq

is2

=1+

2 v 2 Rn inu Gnu + |Yc + Ys |2 + |Yc + Ys |2 2n = 1 + 2 is is Gs Gs

or F =1+

Gnu Rn + [(Gs + Gc )2 + (Xs + Xc )2 ] Gs Gs

(2.5.16)

For a minimum noise factor, Fmin , ∂F Gnu = 0 ⇒ G2s = G2c + = G2opt ∂Gs Rn

(2.5.17)

∂F = 0 ⇒ Xs = −Xc = Xopt ∂Xs

(2.5.18)

and

From (2.5.17), Gnu = Rn (G2opt − G2c )

(2.5.19)

Substituting (2.5.18) and (2.5.19) into (2.5.16), the minimum noise factor is found as Fmin = 1 + 2Rn (Gopt + Gc ) (2.5.20) Using (2.5.18) to (2.5.20), (2.5.16) can be expressed as F = Fmin +

Rn |Ys − Yopt |2 Gs

(2.5.21)

Noise Figure of a Cascaded System Consider a two-port network with gain G1 , noise factor F1 , and equivalent noise temperature Te1 . It is connected in cascade with another two-port network, as shown in Figure 2.14(a). The second two-port network has gain G2 , noise factor

41

NOISE AND DISTORTION

Ni

N1

G1, F1, Te1

Ti

G2, F2, Te2

No

(a) Ni

No

G, F, Te

Ti

(b)

Figure 2.14

Two networks connected in cascade (a) and its equivalent system (b).

F2 , and noise temperature Te2 . Our goal is to find the noise factor F and equivalent noise temperature Te of the overall system illustrated in Figure 2.14(b). Assume that the noise power input to the first two-port network is Ni . Its equivalent noise temperature is Ti . The output noise power of the first system is N1 , whereas it is No after the second system. Hence, N1 = G1 kTi B + G1 kTe1 B

(2.5.22)

and No = G2 N1 + G2 kTe2 B = G1 G2 kB(Ti + Te1 + Te2 /G1 ) = G2 [G1 kB(Ti + Te1 )] + G2 kTe2 B or No = G1 G2 kB(Te + Ti ) = GkB(Te + Ti )

(2.5.23)

Therefore, the noise temperature of a cascaded system is Te = Te1 +

Te2 G1

and from Te1 = (F1 − 1)Ti , Te2 = (F2 − 1)Ti , and Te = (F − 1)Ti , we get F = F1 +

F2 − 1 G1

The equations for Te and F above can be generalized as follows: Te2 Te3 Te4 + + + ··· G1 G1 G2 G1 G2 G3

(2.5.24)

F2 − 1 F3 − 1 F4 − 1 + + +··· G1 G1 G2 G1 G2 G3

(2.5.25)

Te = Te1 + and F = F1 +

42

COMMUNICATION SYSTEMS

Example 2.11 A receiving antenna is connected to an amplifier through a transmission line that has an attenuation of 2 dB. The gain of the amplifier is 15 dB, and its noise temperature is 150 K over a bandwidth of 100 MHz. All the components are at an ambient temperature of 300 K. (a) Find the noise figure of this cascaded system. (b) What would be the noise figure if the amplifier were placed before the transmission line? SOLUTION First we need to determine the noise factor of the transmission line alone. The formulas derived in the preceding section can then provide the noise figures desired for the two cases. Consider a transmission line that is matchterminated at both its ends by resistors R0 , as illustrated in Figure 2.15. Since the entire system is in thermal equilibrium at T (K), noise powers delivered to the transmission line and available at its output are kTB. Mathematically, Po = kTB = GkTB + GN added where Nadded is noise generated by the line as it appears at its input terminals. G is an output-to-input power ratio and B is the bandwidth of the transmission line. Note that input noise is attenuated in a lossy transmission line, but there is noise generated by it as well. Hence, 1 1 Nadded = (1 − G)kTB = − 1 kTB = kTe B G G An expression for the equivalent noise temperature Te of the transmission line is found from it as follows: 1 −1 T Te = G and F =1+

Te =1+ T0

1 T −1 G T0

The gain of the amplifier, Gamp = 15 dB = 101.5 = 31.6228. For the transmission line, 1/G = 100.2 = 1.5849. Hence, the noise factor of the line is 1 + (1.5849 − 1)300/290 = 1.6051. The corresponding noise figure is 2.05 dB. Similarly, the

kTB T R0

G, T, R0

Po = kTB

T R0

Figure 2.15 Lossy transmission line that is matched terminated at its ends.

43

NOISE AND DISTORTION

noise factor of the amplifier is found to be 1 + 150/290 = 1.5172. Its noise figure is 1.81 dB. (a) In this case the noise figure of the cascaded system, Fcascaded , is found as Famp − 1 1.5172 − 1 = 2.4248 = 3.8468 dB = 1.6051 + Gline 1/1.5849

Fcascaded = Fline +

(b) If the amplifier is connected before the line (i.e., the amplifier is placed right at the antenna), Fcascaded = Famp +

Fline − 1 1.6051 − 1 = 1.5363 = 1.8649 dB = 1.5172 + Gamp 31.6228

Note that the amplifier alone has a noise figure of 1.81 dB. Hence, the noisy transmission line connected after it does not alter the noise figure significantly. Example 2.12 Two amplifiers, each with a 20-dB gain, are connected in cascade as shown in Figure 2.16. The noise figure of amplifier A1 is 3 dB, while that of A2 is 5 dB. Calculate the overall gain and noise figure for this arrangement. If the order of two amplifiers is changed in the system, find its resulting noise figure. SOLUTION The noise factors and gains of two amplifiers are F1 = 3 dB = 100.3 = 2 F2 = 5 dB = 100.5 = 3.1623 G1 = G2 = 20 dB = 102 = 100 Therefore, the overall gain and noise figure of the cascaded system is found as follows: G=

P3 P3 P2 = × = 100 × 100 = 10, 000 = 40 dB P1 P2 P1

and F = F1 +

F2 − 1 3.1623 − 1 =2+ = 2.021623 = 3.057 dB G1 100

P1

P2 A1

Figure 2.16

P3 A2

Setup for Example 2.12.

44

COMMUNICATION SYSTEMS

If the order of amplifiers is changed, the overall gain will stay the same. However, the noise figure of the new arrangement will change as follows: F = F2 +

F1 − 1 2−1 = 3.1623 + = 3.1723 = 5.013743 dB G2 100

Minimum Detectable Signal Consider a receiver circuit with gain G over a bandwidth B. Assume that its noise factor is F . PI and Po represent the signal power at its input and output ports, respectively. NI is the input noise power and No is the total noise power at its output, as illustrated in Figure 2.17. Hence, No = kT0 F BG

(2.5.26)

This constitutes the noise floor of the receiver. A signal weaker than this will be lost in noise. No can be expressed in dBW as follows: No (dBW) = 10 log10 kTo + F (dB) + 10 log10 B + G(dB)

(2.5.27)

The minimum detectable signal must have power higher than this. Generally, it is taken as 3 dB above this noise floor. Further, 10 log10 kT0 = 10 log10 (1.38 × 10−23 × 290) ≈ −204 dBW/Hz Hence, the minimum detectable signal PoMDS at the output is PoMDS = −201 + F (dB) + 10 log10 BHz + G(dB)

(2.5.28)

The corresponding signal power PI MDS at its input is PI MDS (dBW) = −201 + F (dB) + 10 log10 BHz

(2.5.29)

Alternatively, PI MDS can be expressed in dBm as follows: PI MDS (dBm) = −111 + F (dB) + 10 log10 BMHz

NI, PI

Receiver

(2.5.30)

No , Po

Figure 2.17 Signals at the two ports of a receiver with a noise figure of F decibels.

45

NOISE AND DISTORTION

Example 2.13 The noise figure of a communication receiver is found as 10 dB at room temperature (290 K). Determine the minimum detectable signal power if (a) B = 1 MHz, (b) B = 1 GHz, (c) B = 10 GHz, and (d) B = 1 kHz. SOLUTION From (2.5.30), PI MDS = −111 + F (dB) + 10 log10 BMHz dBm Hence, (a) PI MDS = −111 + 10 + 10 log10 (1) = −101 dBm = 7.94 × 10−11 mW (b) PI MDS = −111 + 10 + 10 log10 (103 ) = −71 dBm = 7.94 × 10−8 mW (c) PI MDS = −111 + 10 + 10 log10 (104 ) = −61 dBm = 7.94 × 10−7 mW (d) PI MDS = −111 + 10 + 10 log10 (10−3 ) = −131 dBm = 7.94 × 10−14 mW These results show that the receiver can detect a relatively weak signal when its bandwidth is narrow. Intermodulation Distortion The electrical noise of a system determines the minimum signal level that it can detect. On the other hand, the signal will be distorted if its level is too high. This occurs because of the nonlinear characteristics of electrical devices, such as diodes, transistors, and so on. In this section we analyze the distortion characteristics and introduce the associated terminology. Consider the nonlinear system illustrated in Figure 2.18. Assume that its nonlinearity is frequency independent and can be represented by the following power series: (2.5.31) vo = k1 vi + k2 vi2 + k3 vi3 + · · · For simplicity, we assume that the ki are real and the first three terms of this series are sufficient to represent its output signal. Further, it is assumed that the input signal has two different frequency components that can be expressed as follows: (2.5.32) vi = a cos ω1 t + b cos ω2 t Therefore, the corresponding output signal can be written as vo ≈ k1 (a cos ω1 t + b cos ω2 t) + k2 (a cos ω1 t + b cos ω2 t)2 + k3 (a cos ω1 t + b cos ω2 t)3

ni

Figure 2.18

Nonlinear circuit

no

Nonlinear circuits with input signal vi that produces vo at its output.

46

COMMUNICATION SYSTEMS

After simplifying and rearranging it, we get

b2 a2 (1 + cos 2ω1 t) + (1 + cos 2ω2 t) 2 2 ab + {cos(ω1 + ω2 )t + cos(ω1 − ω2 t)} 2 3 3 3 2 a3 4 a cos ω1 t + 2 ab cos ω1 t + 4 cos 3ω1 t 3 2 3 2 + ab cos(ω1 − 2ω2 )t + a b cos(2ω1 − ω2 )t 4 4 (2.5.33) + k3 3 2 3 3 b3 + a b cos ω2 t + b cos ω2 t + cos 3ω2 t 2 4 4 3 2 3 2 + a b cos(2ω1 + ω2 )t + ab cos(ω1 + 2ω2 )t 4 4

vo = k1 (a cos ω1 t + b cos ω2 t) + k2

Therefore, the output signal has several frequency components in its spectrum. Amplitudes of various components are listed in Table 2.6. Figure 2.19 illustrates the input–output characteristic of an amplifier. If the input signal is too low, it may be submerged under the noise. The output power rises linearly above the noise as the input is increased. However, it deviates from the linear characteristic after a certain level of input power. In the linear region, output power can be expressed in dBm as follows: Pout (dBm) = Pin (dBm) + G(dB) The input power for which output deviates by 1 dB below its linear characteristic is known as 1-dB compression point. In this figure, it occurs at an input power

1 dB

Po1

Pout(dBm)

Noise level 1- dB compression point

Dynamic range PI MDS

Figure 2.19

Pin(dBm)

PD

Gain characteristics of an amplifier.

NOISE AND DISTORTION

47

TABLE 2.6 Amplitudes of Various Harmonics in the Output Harmonic Components ω1 ω2 ω1 − ω2 ω1 + ω2 2ω1 2ω2 3ω1 3ω2 2ω1 − ω2 ω1 − 2ω2 2ω1 + ω2 ω1 + 2ω2

Amplitude 3 3 3 2 a + ab k1 a + k3 4 2 3 3 3 2 b + a b k1 b + k3 4 2 ab k2 2 ab k2 2 a2 k2 2 b2 k2 2 a3 k3 4 b3 k3 4 3 k3 a 2 b 4 3 k3 ab2 4 3 k3 a 2 b 4 3 k3 ab2 4

of PD (dBm) that produces an output of Po1 (dBm). From the relation above, we find that Po1 (dBm) + 1 = PD (dBm) + G(dB) or PD (dBm) = Po1 (dBm) + 1 − G(dB)

(2.5.34)

The difference between the input power at 1-dB compression point and the minimum detectable signal defines the dynamic range (DR). Hence, DR = PD (dBm) − PI MDS From (2.5.30) and (2.5.34), we find that DR = Po1 (dBm) + 112 − G(dB) − F (dB) − 10 log10 BMHz

(2.5.35)

48

COMMUNICATION SYSTEMS

Gain Compression Nonlinear characteristics of the circuit (amplifier, mixer, etc.) compress its gain. If there is only one input signal [i.e., b is zero in (2.5.32)], the amplitude a1 of cos ω1 t in its output is found from Table 2.6 to be a1 = k1 a + 34 k3 a 3

(2.5.36)

The first term of a1 represents the linear (ideal) case, while its second term results from the nonlinearity. At the 1-dB compression point, k1 a + 34 k3 a 3 k1 a + 34 k3 a 3 dB = k1 a|dB − 1 → 20 log k1 a = −1 = −20 log (1.122) or

k1 + 34 k3 a 2 k1 + 34 k3 a 2 × 1.122 = 0 → × 1.122 = 1 20 log k1 k1

Therefore, k3 = −0.145

k1 a2

(2.5.37)

This indicates that k3 is a negative constant for a positive k1 , or vice versa. Therefore, it tends to reduce a1 , resulting in a lower gain. The single-tone gain compression factor may be defined as follows: A1C =

a1 3k3 2 a =1+ k1 a 4k1

(2.5.38)

Let us now consider the case when both of the input signals are present in (2.5.32). The amplitude of cos ω1 t in the output now becomes k1 a + k3 34 a 3 + 32 ab2 . If b is large compared with a, the term with k3 may dominate (undesired) over the first (desired) one. Since k3 and k1 have opposite signs, the desired output k1 a may be blocked completely for a certain value of b. Second Harmonic Distortion Second harmonic distortion occurs due to k2 . If b is zero, the amplitude of the a2 second harmonic will be k2 . Since power is proportional to the square of 2 the voltage, the desired term in the output can be expressed as a 2 = 20 log10 a + C1 P1 = 10 log10 ζk1 2

(2.5.39)

49

NOISE AND DISTORTION

where ζ is the proportionality constant and C1 = 20 log

ζk1 2

Similarly, power in the second harmonic component can be expressed as P2 = 10 log10

a2 ςk2 2

2 = 40 log10 a + C2

(2.5.40)

and the input power is a 2 = 20 log10 a + C3 Pin = 10 log10 ς 2

(2.5.41)

Proportionality constants ς and k2 are embedded in C2 and C3 . From (2.5.39)– (2.5.41) we find that P1 = Pin + D1

(2.5.42)

P2 = 2Pin + D2

(2.5.43)

and

where D1 and D2 replace C1 − C3 and C2 − 2C3 , respectively. Equations (2.5.42) and (2.5.43) indicate that both the fundamental and second harmonic signals in the output are linearly related with input power. However, the second harmonic power increases at twice the rate of the fundamental (the desired) component.

Intermodulation Distortion Ratio From Table 2.6 we find that the cubic term produces intermodulation frequencies 2ω1 ± ω2 and 2ω2 ± ω1 . If ω1 and ω2 are very close, 2ω1 + ω2 and 2ω2 + ω1 will be far away from the desired signals, and therefore, these can be filtered out easily. However, the other two terms, 2ω1 − ω2 and 2ω2 − ω1 , will be so close to ω1 and ω2 that these components may be within the passband of the system. It will distort the output. This characteristic of a nonlinear circuit is specified via the intermodulation distortion. It is obtained after dividing the amplitude of one of the intermodulation terms by the desired output signal. For an input signal with both ω1 and ω2 (i.e., a two-tone input), the intermodulation distortion ratio (IMR) may be found as IMR =

3 k a2b 4 3

k1 a

=

3k3 ab 4k1

(2.5.44)

50

COMMUNICATION SYSTEMS

Intercept Point Since power is proportional to the square of the voltage, the intermodulation distortion power may be defined as 3 PIMD = ς

ka 4 3

2

b

2 (2.5.45)

2

where ς is the proportionality constant. If the two input signals are equal in amplitude, a = b and the expression for intermodulation distortion power simplifies to 3 2 k a3 4 3 PIMD = ς (2.5.46) 2 Similarly, the power in one of the input signal components can be expressed as Pin = ς

(a)2 2

Therefore, the intermodulation distortion power can be expressed as PIMD = αPin3 where α is another constant. The ratio of intermodulation distortion power (PIMD ) to the desired output power Po (Po = k12 Pin ) for the case where two input signal amplitudes are the same is known as the intermodulation distortion ratio. It is found to be PIMR =

PIMD = α1 Pin2 Po

(2.5.47)

Note that PIMD increases as the cube of input power and the desired signal power Po is linearly related with Pin . Hence, PIMD increases three times as fast as Po on a log-log plot (or both of them are expressed in dBm before displaying on a linear graph). In other words, for a change of 1 dBm in Pin , Po changes by 1 dBm, whereas PIMD changes by 3 dBm. The value of the input power for which PIMD is equal to Po is referred to as the intercept point (IP). It is also referred to as IIP3 and the corresponding output power as OIP3 . Hence, PIMR is unity at the intercept point. If PIP is input power at IP, PIMR = 1 = α1 PIP2 ⇒ α1 =

1 PIP2

(2.5.48)

Therefore, the intermodulation distortion ratio PIMR is related to PIP as PIMR =

Pin PIP

2 (2.5.49)

51

NOISE AND DISTORTION

Example 2.14 The intercept point in the transfer characteristic of a nonlinear system is found to be at 25 dBm. If a −15-dBm signal is applied to this system, find the intermodulation ratio. SOLUTION PIMR =

Pin PIP

2 → PIMR (dB) = 2[Pin (dBm) − PIP (dBm)] = 2(−15 − 25) = −80 dB

Dynamic Range As mentioned earlier, noise at one end and distortion at the other limit the range of detectable signals of a system. The amount of distortion that can be tolerated depends somewhat on the type of application. If we set the upper limit that a system can detect as the signal level at which intermodulation distortion is equal to the minimum detectable signal, we can formulate an expression for its dynamic range. Thus, the ratio of the signal power that causes distortion power (in one frequency component) to be equal to the noise floor and the minimum detectable signal is the dynamic range (DR) of the system (amplifier, mixer, or receiver). It is also known as the spurious-free dynamic range (SFDR). Since the ideal power output Po is linearly related to input as Po = k12 Pin

(2.5.50)

the distortion power with reference to input can be expressed as Pdi =

PIMD k12

(2.5.51)

Therefore, PIMR =

k 2 Pdi PIMD Pdi = 12 = = Po Pin k1 Pin

Pin PIP

2 (2.5.52)

If Pdi = Nf (noise floor at the input), Nf = Pin

Pin PIP

2 ⇒ Pin3 = PIP2 Nf ⇒ Pin = (PIP2 Nf )1/3

(2.5.53)

and the dynamic range is (P 2 Nf )1/3 = DR = IP Nf

PIP Nf

2/3 (2.5.54)

or DR(dB) = 23 [PIP (dBm) − Nf (dBm)]

(2.5.55)

52

COMMUNICATION SYSTEMS

Example 2.15 A receiver is operating at 900 MHz with its bandwidth at 500 kHz and the noise figure at 8 dB. If its input impedance is 50 and IP is 10 dBm, find its dynamic range. SOLUTION Nf = kT0 BF ⇒ Nf (dBm) = 10 log10 (1.38 × 10−23 × 290 × 500 × 103 × 103 ) + 8 = −108.99 dBm Therefore, DR = 23 (10 + 108.99) = 79.32 dB

SUGGESTED READING Balanis, C. A. , Antenna Theory. New York: Wiley, 1997. Cheah, J. Y. C., Introduction to wireless communications applications and circuit design, in L. E. Larsen (ed.), RF and Microwave Circuit Design for Wireless Communications. Boston: Artech House, 1996. Razavi, B., RF Microelectronics. Upper Saddle River, NJ: Prentice Hall, 1998. Schiller, J. H., Mobile Communication. Reading, MA: Addison-Wesley, 2000. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1997. Stutzman, W. L., and G. A. Thiele, Antenna Theory and Design. New York: Wiley, 1998.

PROBLEMS 2.1. (a) What is the gain in decibels of an amplifier with a power gain of 4? (b) What is the power gain ratio of a 5-dB amplifier? (c) Express 2 kW of power in terms of dBm and dBw. 2.2. The maximum radiation intensity of a 90% efficient antenna is 200 mW per unit solid angle. Find the directivity and gain (dimensionless and in dB) when (a) the input power is 40π milliwatts and (b) the radiated power is 40π milliwatts. 2.3. The normalized far-zone field pattern of an antenna is given by sin θ cos2 φ E= 0

0 θ π and 0 φ π/2, 3π/2 φ 2π elsewhere

Find the directivity of this antenna.

PROBLEMS

53

2.4. The radiation intensity pattern of an antenna is given by U (θ, φ) =

2.5.

2.6.

2.7.

2.8.

2.9.

10 sin θ cos2 φ 0 θ π and 0 φ π/2, 3π/2 φ 2π 0 elsewhere

Find the directivity of this antenna. For an X-band (8.2 to 12.4 GHz) rectangular antenna with aperture dimensions of 5.5 and 7.4 cm, find its maximum effective aperture (in cm2 ) when its gain is (a) 14.8 dB at 8.2 GHz, (b) 16.5 dB at 10.3 GHz, and (c) 18.0 dB at 12.4 GHz. Transmitting and receiving antennas operating at 1 GHz with gains of 20 and 15 dB, respectively, are separated by a distance of 1 km. Find the maximum power delivered to the load when the input power is 150 W, assuming that the antennas are polarization matched. An antenna with a total loss resistance of 1 is connected to a generator whose internal impedance is 50 + j 25. Assuming that the peak voltage of the generator is 2 V and the impedance of antenna is 74 + j 42.5, find the power (a) supplied by the source (real power), (b) radiated by the antenna, and (c) dissipated by the antenna. The antenna connected to a radio receiver induces 9 µV of root-meansquare voltage into its input impedance, which is 50 . Calculate the input power in watts, dBm, and dBW. If the signal is amplified by 127 dB before it is fed to an 8- speaker, find the output power. The electric field radiated by a rectangular aperture, mounted on an infinite ground plane with z perpendicular to the aperture, is given by E = θˆ cos φ − φˆ sin φ cos θ f (r, θ, φ)

where f (r, θ, φ) is a scalar function that describes the field variation of the antenna. Assuming that the receiving antenna is linearly polarized along the x-axis, find the polarization loss factor. 2.10. A radar receiver has a sensitivity of 10−12 W. If the radar antenna’s effective aperture is 1 m2 and the wavelength is 10 cm, find the transmitter power required to detect an object at a distance of 3 km with a radar cross section of 5 m2 . 2.11. Two space vehicles are separated by 108 m. Each has an antenna with D = 1000 operating at 2.5 GHz. If vehicle A’s receiver requires 1 pW for a 20-dB signal-to-noise ratio, what transmitter power is required on vehicle B to achieve this signal-to-noise ratio? 2.12. (a) Design an Earth-based radar system that receives 10−15 W of peak echo power from Venus. It is to operate at 10 GHz with a single antenna to be used for both transmitting and receiving. Specify the effective aperture of the antenna and the peak transmitter power. Assume that the Earth to

54

COMMUNICATION SYSTEMS

Venus distance is 3 light-minutes; the diameter of Venus is 13 × 106 m; and the radar cross section of Venus is 15% of its physical cross section. (b) If the system of part (a) is used to observe the moon, determine the power received. Assume that the moon diameter is 3.5 × 106 m, its radar cross section is 15% of the physical cross section, and the Earthto-moon distance is 1.2 light-seconds. 2.13. Consider an imaging satellite that sends close-up pictures of the planet Neptune. It uses a 10-W transmitter operating at 18 GHz and a 2.5-mdiameter parabolic dish antenna. What Earth-station system temperature is required to provide a signal-to-noise ratio of 3 dB for reception of a picture with 3 × 106 pixels (picture elements) in 2 min if the Earth-station antenna diameter is 75 m? Assume aperture efficiencies of 70% and the Earth–Neptune distance as 4 light-hours. One pixel is equal to one bit, and two bits per second has a bandwidth of 1 Hz. 2.14. A Doppler radar operating at 10 GHz employs an antenna with its gain at 20 dB. If the signal reflected back from a target is a uniform plane wave with a power density of 10−8 W/m2 at the antenna, find the power delivered to a matched receiver. Also, if this radar is intended to detect target velocities ranging from 90 to 130 km/h, find the required passband of the Doppler filter. 2.15. A satellite transmitter is 36,500 km away from the receiver. It radiates a power of 10 W through an antenna of 19 dB gain toward the receiver. The satellite operates at a frequency of 11.5 GHz. If the receiving antenna has a gain of 60.5 dB, find the received power in dBm. 2.16. Consider the WLAN receiver front end shown in Figure P2.16, in which the bandpass filter has a bandwidth of 250 MHz centered at 5.4 GHz. If the system is operating at room temperature (290 K), calculate (a) the overall noise figure in decibels and (b) the minimum detectable signal in milliwatts. (c) Suggest if there is a better arrangement to increase the signal to noise ratio at the output.

Pin = −90 dBm

BPF

Pout

G = −1.5 dB

G = 10 dB F = 2 dB

G = 15 dB F = 2 dB

Figure P2.16

2.17. Various units of a 4-GHz receiver have the gains and noise temperatures indicated in Figure P2.17. (a) If the transmission line has a loss of 2 dB and the system is at an ambient temperature of 300 K, find the noise figure (in decibels) of this receiver.

55

PROBLEMS

RF amplifier

Mixer

IF amplifier

TRF = 50 K GRF = 50 dB

Tm = 500 K Gm = −10 dB

TIF = 1000 K GIF = 30 dB

Transmission line

Figure P2.17

(b) Suggest a better receiver architecture (if it exists) that improves the noise figure. Also, find the noise temperature of the new arrangement. 2.18. Two receivers for a design trade-off study are shown in Figure P2.18(a) and (b).

Mixer

IF amplifier

12−14 GHz

IF amplifier 2− 4 GHz

BPF

10 GHz

F = 6 dB G = −6 dB

F = 2 dB G = −2 dB

F = 5 dB G = 10 dB

F = 2 dB G = 15 dB

LO (a)

12−14 GHz

RF amplifier

F = 5 dB G = 10 dB

Mixer

10 GHz

IF amplifier F = 6 dB G = −6 dB

BPF F = 2 dB G = −2 dB

2 −4 GHz F = 4 dB G = 15 dB

LO (b)

Figure P2.18

(a) Calculate the noise figure for the two systems. (b) Calculate the overall gain for the two systems. (c) Calculate the minimum detectable signal at the input for the two systems. (d) If the output power at the 1-dB compression point is 10 mW for both systems, calculate the dynamic range for the two systems. 2.19. Calculate the input minimum detectable signal in milliwatts at room temperature for a receiver with (a) BW = 1 GHz, F = 5 dB and (b) BW = 100 MHz, F = 10 dB. 2.20. (a) What is the root-mean-square (rms) noise voltage produced in a 10-k resistance when its temperature is 45◦ C and the effective bandwidth is 100 MHz?

56

COMMUNICATION SYSTEMS

(b) A 50-k resistance is connected in parallel with the 10-k resistance of part (a). What is the resulting rms noise voltage? 2.21. The intercept point in the transfer characteristic of a nonlinear system is found to be 33 dBm. If a −18-dBm signal is applied to this system, find the intermodulation ratio. 2.22. A receiver is operating at 2455 MHz with its bandwidth at 500 kHz and the noise figure at 15 dB. If its input impedance is 50 and IP is 18 dBm, find its dynamic range. 2.23. An amplifier has a gain of 6 dB. Its intermodulation ratio is found to be −40 dB at an input power level of −4 dBm. Calculate the values of its input and output intercept points. What input signal level is needed if we want an intermodulation ratio of −50 dB?

3 TRANSMISSION LINES

Transmission lines are needed for connecting various circuit elements and systems. Open wire and coaxial lines are commonly used for circuits operating at low frequencies. On the other hand, coaxial line, strip line, microstrip line, and waveguides are employed at radio and microwave frequencies. Generally, lowfrequency signal characteristics are not affected as the signal propagates through the line. However, radio frequency and microwave signals are affected significantly because of the fact that the circuit size is comparable to the wavelength. A comprehensive understanding of signal propagation requires the analysis of electromagnetic fields in a given line. On the other hand, a generalized formulation can be obtained using circuit concepts on the basis of line parameters. This chapter begins with an introduction to line parameters and a distributed model of the transmission line. Solutions to the transmission line equation are then constructed to explain the behavior of the propagating signal. This is followed by discussions on sending-end impedance, reflection coefficient, return loss, and insertion loss. A quarter-wave impedance transformer is also presented, along with a few examples of matching resistive loads. Impedance measurement using the voltage standing wave ratio is then discussed. Finally, the Smith chart is introduced to facilitate graphical analysis and design of transmission line circuits. 3.1 DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES Any transmission line can be represented by a distributed electrical network (Figure 3.1) comprised of series inductors and resistors and shunt capacitors and Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

57

58

TRANSMISSION LINES R ∆z L ∆z

R ∆z

R ∆z

L ∆z C ∆z

G ∆z

L ∆z C ∆z

G ∆z

C ∆z

G ∆z

∆z

Figure 3.1 Distributed network model of a transmission line.

resistors. These distributed elements are defined as follows: L = inductance per unit length (H/m) R = resistance per unit length (/m) C = capacitance per unit length (F/m) G = conductance per unit length (S/m) L, R, C, and G, the line parameters, are determined theoretically by electromagnetic field analysis of a transmission line. These parameters are influenced by their cross-section geometry and the electrical characteristics of their constituents. For example, if a line is made up of an ideal dielectric and a perfect conductor, its R and G values will be zero. If it is a coaxial cable with inner and outer radii a and b, respectively, as shown in Figure 3.2, then C=

55.63εr ln(b/a)

pF/m

(3.1.1)

and L = 200 ln(b/a)

nH/m

a

εr

b

Figure 3.2 Coaxial line geometry.

(3.1.2)

59

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

where εr is the dielectric constant of the material between two coaxial conductors of the line. If the coaxial line has small losses due to an imperfect conductor and insulator, its resistance and conductance parameters can be calculated as follows: 1 1 fGHz R ≈ 10 + /m (3.1.3) a b σ and G=

0.3495εr fGHz tan δ ln(b/a)

S/m

(3.1.4)

where tanδ is the loss tangent of the dielectric material, σ the conductivity (in S/m) of the conductors, and fGHz is the signal frequency in gigahertz. Characteristic Impedance of a Transmission Line Consider a transmission line that extends to infinity, as shown in Figure 3.3. The voltages and currents at several points on the line are as indicated. When a voltage is divided by the current through that point, the ratio is found to remain constant. This ratio is called the characteristic impedance of the transmission line. Mathematically, characteristic impedance = Z0 = V1 /I1 = V2 /I2 = V3 /I3 = · · · = Vn /In In actual electrical circuits, the length of the transmission lines is always finite. Hence, it seems that the characteristic impedance has no significance in the real world. However, that is not the case. When the line extends to infinity, an electrical signal continues propagating in the forward direction without reflection. On the other hand, it may be reflected back by the load that terminates a transmission line of finite length. If one varies this termination, the strength of the reflected signal changes. When the transmission line is terminated by a load impedance that absorbs all the incident signal, the voltage source sees an infinite electrical length. The voltage-to-current ratio at any point on this line is a constant equal

I1

I2

V1

V2

I3

In

V3

Vn

Figure 3.3 Infinitely long transmission line and voltage source.

∞

60

TRANSMISSION LINES

to the terminating impedance. In other words, there is a unique impedance for every transmission line that does not produce an echo signal when the line is terminated by it. The terminating impedance that does not produce an echo on the line is equal to its characteristic impedance. If Z = R + j ωL = impedance per unit length Y = G + j ωC = admittance per unit length then using the definition of characteristic impedance and the distributed model shown in Figure 3.1, we can write Z0 = =

(Z0 + Z z)(1/Y z) Z0 + Z z + 1/Y z Z0 + Z z ⇒ Z0 Y (Z0 + Z z) = Z 1 + Y z(Z0 + Z z)

For z → 0,

Z0 =

Z = Y

R + j ωL G + j ωC

(3.1.5)

Some special cases are as follows: √ 1. For a dc signal, Z0 dc = R/G. √ 2. For ω → ∞, ωL R and ωC G; therefore, Z0(ω→large) = L/C. √ 3. For a lossless line, R → 0 and G → 0; therefore, Z0 = L/C. Thus, a lossless semirigid coaxial line with 2a = 0.036 in., 2b = 0.119 in., and an εr value of 2.1(Teflon-filled) will have C = 97.71 pF/m and L = 239.12 nH/m. Its characteristic impedance will be 49.5 . Since the conductivity of copper is 5.8 × 107 S/m and the loss tangent of Teflon is 0.00015, Z = 3.74 + j 1.5 × 103 /m and Y = 0.092 + j 613.92 mS/m at 1 GHz. The corresponding characteristic impedance is 49.5 − j 0.058 , which is very close to the approximate value of 49.5 . Example 3.1 Calculate the equivalent impedance and admittance of a 1-mlong line that is operating at 1.6 GHz. The line parameters are L = 0.002 µ H/m, C = 0.012 pF/m, R = 0.015 /m, and G = 0.1 mS/m. What is the characteristic impedance of this line? SOLUTION Z = R + j ωL = 0.015 + j 2π × 1.6 × 109 × 0.002 × 10−6 /m = 0.015 + j 20.11 /m

61

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

Y = G + j ωC = 0.0001 + j 2π × 1.6 × 109 × 0.012 × 10−12 S/m Z0 =

= 0.1 + j 0.1206 mS/m Z = 337.02 + j 121.38 Y

Transmission Line Equations Consider the equivalent distributed circuit of a transmission line that is terminated by a load impedance ZL , as shown in Figure 3.4. The line is excited by a voltage source v(t) with its internal impedance ZS . We apply Kirchhoff’s voltage and current laws over a small length, z, of this line as follows: For the loop, v(z, t) = L z or

∂i(z, t) + R z i(z, t) + v(z + z, t) ∂t

∂i(z, t) v(z + z, t) − v(z, t) = −Ri(z, t) − L z ∂t

Under the limit z → 0, the equation above reduces to ∂i(z, t) ∂v(z, t) = − R × i(z, t) + L ∂z ∂t

(3.1.6)

Similarly, at node A, i(z, t) = i(z + z, t) + G z v(z + z, t) + C z

∂v(z + z, t) ∂t

or ∂v(z + z, t) i(z + z, t) − i(z, t) = − G × v(z + z, t) + C z ∂t

ZS

v (z + ∆z, t) i(z + ∆z, t) A

v(z,t) L ∆z

v(t)

R ∆z C ∆z

z

Figure 3.4

i(z,t) L ∆z R ∆z G ∆z

C ∆z

L ∆z G ∆z

R ∆z C ∆z

∆z

Distributed circuit model of a transmission line.

G ∆z

ZL

62

TRANSMISSION LINES

Again, under the limit z → 0, it reduces to ∂v(z, t) ∂i(z, t) = −Gv(z, t) − C ∂z ∂t

(3.1.7)

Now, from equations (3.1.6) and (3.1.7), v(z, t) or i(z, t) can be eliminated to formulate the following: ∂ 2 v(z, t) ∂v(z, t) ∂ 2 v(z, t) = RGv(z, t) + (RC + LG) + LC ∂z2 ∂t ∂t 2

(3.1.8)

∂i(z, t) ∂ 2 i(z, t) ∂ 2 i(z, t) = RGi(z, t) + (RC + LG) + LC ∂z2 ∂t ∂t 2

(3.1.9)

and

Some special cases are as follows: 1. For a lossless line, R and G will be zero, and these equations reduce to well-known homogeneous scalar wave equations, ∂ 2 v(z, t) ∂ 2 v(z, t) = LC ∂z2 ∂t 2

(3.1.10)

∂ 2 i(z, t) ∂ 2 i(z, t) = LC ∂z2 ∂t 2

(3.1.11)

and

√ It is to be noted that the velocity of these waves is 1/ LC. 2. If the source is sinusoidal with time (i.e., time-harmonic), we can switch to phasor voltages and currents. In that case, equations (3.1.8) and (3.1.9) can be simplified as follows: d 2 V (z) = ZY V (z) = γ2 V (z) dz2

(3.1.12)

d 2 I (z) = ZY I (z) = γ2 I (z) dz2

(3.1.13)

and

where V (z) and I (z) are phasor quantities; Z and Y are impedance per unit length √ and admittance per unit length, respectively, as defined earlier. γ = ZY = α + j β is known as the propagation constant of the line. α and β are called the attenuation constant and the phase constant, respectively. Equations (3.1.12) and (3.1.13) are referred to as homogeneous Helmholtz equations.

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

63

Solution of Helmholtz Equations Note that both differential equations have the same general format. Therefore, we consider the solution to the following generic equation here. Expressions for voltage and current on the line can be constructed on the basis of that. d 2 f (z) − γ2 f (z) = 0 dz2

(3.1.14)

Assume that f (z) = Ceκz , where C and κ are arbitrary constants. Substituting it into (3.1.14), we find that κ = ±γ. Therefore, a complete solution to this equation may be written as follows: f (z) = C1 e−γz + C2 eγz

(3.1.15)

where C1 and C2 are integration constants that are evaluated through the boundary conditions. Hence, complete solutions to equations (3.1.12) and (3.1.13) can be written as follows: V (z) = Vin e−γz + Vref eγz

(3.1.16)

I (z) = Iin e−γz + Iref eγz

(3.1.17)

and

where Vin , Vref , Iin , and Iref are integration constants that may be complex, in general. These constants can be evaluated from the known values of voltages and currents at two different locations on the transmission line. If we express the first two of these constants in polar form as Vin = vin ej φ

and Vref = vref ej ϕ

the line voltage, in time domain, can be evaluated as follows: v(z, t) = Re[V (z)ej ωt ] = Re[Vin e−(α+j β)z ej ωt + Vref e+(α+j β)z ej ωt ] or v(z, t) = vin e−αz cos(ωt − βz + φ) + vref e+αz cos(ωt + βz + ϕ) (3.1.18) At this point, it is important to analyze and understand the behavior of each term on the right-hand side of this equation. At a given time, the first term changes sinusoidally with distance, z, while its amplitude decreases exponentially. It is illustrated in Figure 3.5(a). On the other hand, the amplitude of the second sinusoidal term increases exponentially. It is shown in Figure 3.5(b). Further, the argument of the cosine function decreases with distance in the former case and

TRANSMISSION LINES 1.2

150.0

0.8

100.0

0.4

50.0

f2 (Z)

f1 (Z)

64

0.0 −0.4 −0.8 0.0

−50.0

λ 1.0

0.0

2.0

3.0

4.0

5.0

−100.0 0.0

λ 1.0

Z (a)

Figure 3.5

2.0

3.0

4.0

5.0

Z (b)

Behavior of two solutions to the Helmholtz equation with distance.

increases in the latter case. When a signal is propagating away from the source along +z-axis, its phase should be delayed. Further, if it is propagating in a lossy medium, its amplitude should decrease with distance z. Thus, the first term on the right-hand side of equation (3.1.16) represents a wave traveling along the +z-axis (an incident or outgoing wave). Similarly, the second term represents a wave traveling in the opposite direction (a reflected or incoming wave). This analysis is also applied to equation (3.1.17). Note that Iref is reflected current that will be 180◦ out of phase with incident current Iin . Hence, Vref Vin =− = Z0 Iin Iref and therefore equation (3.1.17) may be written as follows: I (z) =

Vin −γz Vref γz e − e Z0 Z0

(3.1.19)

Incident and reflected waves change sinusoidally with both space and time. Time duration over which the phase angle of a wave goes through a change of 360◦ (2π radians) is known as its time period. The inverse of the time period in seconds is the signal frequency in hertz. Similarly, the distance over which the phase angle of the wave changes by 360◦ (2π radians) is known as its wavelength (λ). Therefore, the phase constant β is equal to 2π divided by the wavelength in meters. Phase and Group Velocities The velocity with which the phase of a time-harmonic signal moves is known as its phase velocity. In other words, if we tag a phase point of the sinusoidal

DISTRIBUTED CIRCUIT ANALYSIS OF TRANSMISSION LINES

65

wave and monitor its velocity, we obtain the phase velocity, vp , of this wave. Mathematically, ω vp = β A transmission line has no dispersion if the phase velocity of a propagating signal is independent of frequency. Hence, a graphical plot of ω versus β will be a straight line passing through the origin. This type of plot is called the dispersion diagram of a transmission line. An information-carrying signal is composed of many sinusoidal waves. If the line is dispersive, each of these harmonics will travel at a different velocity. Therefore, the information will be distorted at the receiving end. The velocity with which a group of waves travels is called the group velocity, vg . It is equal to the slope of the dispersion curve of the transmission line. Consider two sinusoidal signals with angular frequencies ω + δω and ω − δω, respectively. Assume that these waves of equal amplitudes are propagating in the z-direction with corresponding phase constants β + δβ and β − δβ. The resulting wave can be found as follows: f (z, t) = Re[Aej [(ω+δω)t−(β+δβ)z] + Aej [(ω−δω)t−(β−δβ)z] ] = 2A cos(δωt − δβz) cos(ωt − βz) Hence, the resulting wave, f (z, t), is amplitude modulated. The envelope of this signal moves with the group velocity, vg =

δω δβ

Example 3.2 A signal generator has an internal resistance of 50 and an opencircuit voltage v(t) = 3 cos(2π × 108 t) V. It is connected to a 75- lossless transmission line that is 4 m long and terminated by a matched load at the other end. If the signal propagation velocity on this line is 2.5 × 108 m/s, find the instantaneous voltage and current at an arbitrary location on the line. SOLUTION Since the transmission line is terminated by a load that is equal to its characteristic impedance, there will be no echo signal. Further, an equivalent circuit at its input end may be drawn as shown in Figure 3.6. Using the voltagedivision rule and Ohm’s law, incident voltage and current can be determined as follows: 75 ◦ ◦ 3 0 = 1.8 0 V 50 + 75 3 0◦ ◦ incident current at the input end, Iin (z = 0) = = 0.024 0 A 50 + 75

incident voltage at the input end, Vin (z = 0) =

66

TRANSMISSION LINES 50 Ω

50 Ω

4m

75 Ω

3V∠0°

75 Ω

75 Ω 3V∠0°

z

Figure 3.6

and β=

Circuit arrangement for Example 3.2.

2π × 108 ω = = 0.8π rad/m vp 2.5 × 108

Therefore, V (z) = 1.8e−j 0.8πz V and I (z) = 0.024e−j 0.8πz A. Hence, v(z, t) = 1.8 cos(2π × 108 t − 0.8πz) V and i(z, t) = 0.024 cos(2π × 108 t − 0.8πz) A Example 3.3 The parameters of a transmission line are R = 2 /m, G = 0.5 mS/m, L = 8 nH/m, and C = 0.23 pF/m. If the signal frequency is 1 GHz, calculate its characteristic impedance (Z0 ) and the propagation constant (γ). SOLUTION R + j ωL 2 + j 2π × 109 × 8 × 10−9 = Z0 = G + j ωC 0.5 × 10−3 + j 2π × 109 × 0.23 × 10−12 2 + j50.2655 50.31 1.531 rad = = −3 −3 0.5 × 10 + j 1.4451 × 10 15.29 × 10−4 1.2377 rad ◦

= 181.39 8.4 = 179.44 + j 26.51 and γ=

√ ZY = (50.31 1.531 rad) × (15.29 × 10−4 1.2377 rad) ◦

= 0.2774 79.31 m−1 = 0.0514 + j 0.2726 m−1 = α + j β Therefore, α = 0.0514 Np/m and β = 0.2726 rad/m. Example 3.4 Two antennas are connected through a quarter-wavelength-long lossless transmission line, as shown in Figure 3.7. However, the characteristic

67

SENDING-END IMPEDANCE

jX Z0 = ?

B

80 + j35 Ω

A

50 Ω 56 + j28 Ω

(λ/4)

z

Figure 3.7

Circuit arrangement for Example 3.4.

impedance of this line is unknown. The array is excited through a 50- line. Antenna A has an impedance of 80 + j 35 and antenna B has 56 + j 28 . Currents (peak values) through these antennas are found to be 1.5 0◦ and 1.5 90◦ A, respectively. Determine the characteristic impedance of the line connecting these two antennas and the value of a reactance connected in series with antenna B. SOLUTION Assume that Vin and Vref are the incident and reflected phasor voltages, respectively, at antenna A. Therefore, the current, IA , through this antenna is IA =

Vin − Vref ◦ ◦ = 1.5 0 A ⇒ Vin − Vref = Z0 IA = Z0 1.5 0 V Z0

Since the connecting transmission line is a quarter-wavelength long, the incident and reflected voltages across the transmission line at the location of B will be j Vin and −j Vref , respectively. Therefore, the total voltage, VTBX , appearing across the antenna B and the reactance jX combined will be equal to j (Vin − Vref ). ◦

VTBX = j (Vin − Vref ) = jZ0 IA = j 1.5Z0 = 1.5Z0 90 Also, Ohm’s law can be used to find this voltage as follows: ◦

VTBX = (56 + j 28 + j X)1.5 90

Therefore, X = −28 and Z0 = 56 . Note that the unknown characteristic impedance is a real quantity because the transmission line is lossless. 3.2 SENDING-END IMPEDANCE Consider a transmission line of length l and characteristic impedance Z0 . It is terminated by a load impedance ZL , as shown in Figure 3.8. Assume that the

68

TRANSMISSION LINES Iin Iref

Zin

z=0

Figure 3.8

Load ZL

Z0

V in + Vref

z=l

Transmission line terminated by a load impedance.

incident and reflected voltages at its input (z = 0) are Vin and Vref , respectively. The corresponding currents are represented by Iin and Iref . If V (z) represents the total phasor voltage at point z on the line and I (z) is the total current at that point, V (z) = Vin e−γz + Vref eγz

(3.2.1)

I (z) = Iin e−γz + Iref eγz

(3.2.2)

and

where Vin , Vref , Iin , and Iref are incident voltage, reflected voltage, incident current, and reflected current at z = 0, respectively. Impedance at the input of this transmission line, Zin , can be found after dividing total voltage by the total current at z = 0. Thus, Zin =

V (z = 0) Vin + Vref Vin + Vref Vin + Vref = = Z0 = I (z = 0) Iin + Iref Vin /Z0 − Vref /Z0 Vin − Vref

or Zin = Z0

1 + Vref /Vin 1 + 0 = Z0 1 − Vref /Vin 1 − 0

(3.2.3)

where 0 = ρej φ is known as the input reflection coefficient. Further, Zin = Z0

Zin 1 + 0 1 + 0 ⇒ = Z in = 1 − 0 Z0 1 − 0

where Z in is called the normalized input impedance. Similarly, voltage and current at z = l are related through load impedance as follows: ZL =

V (z = l) Vin e−γl + Vref e+γl = I (z = l) Iin e−γl + Iref e+γl

= Z0

Vin e−γl + Vref e+γl e−γl + 0 e+γl = Z 0 −γl Vin e−γl − Vref e+γl e − 0 e+γl

69

SENDING-END IMPEDANCE

Therefore, ZL =

e−γl + 0 eγl ZL − 1 −2γl ⇒ 0 = e e−γl − 0 e+γl ZL + 1

(3.2.4)

and equation (3.2.3) can be written as follows: Zin =

1 + [(ZL − 1)/(ZL + 1)]e−2γl 1 − [(ZL − 1)/(ZL + 1)]e−2γl

=

ZL (1 + e−2γl ) + (1 − e−2γl ) ZL (1 − e−2γl ) + (1 + e−2γl )

since 1 − e−2γl e+γl − e−γl sinh γl = tanh γl = +γl = −2γl 1+e e + e−γl cosh γl Z in =

Z L + tanh γl 1 + Z L tanh γl

or Zin = Z0

ZL + Z0 tanh γl Z0 + ZL tanh γl

(3.2.5)

For a lossless line, γ = α + j β = j β, and therefore tanh γl = tanh j βl = j tan βl. Hence, equation (3.2.5) simplifies as follows: Zin = Z0

ZL + j Z0 tan βl Z0 + j ZL tan βl

(3.2.6)

It is to be noted from this equation that Zin repeats periodically every one-half wavelength on the transmission line. In other words, the input impedance on a lossless transmission line will be the same at points d ± nλ/2, where n is an integer, due to the fact that βl =

2π nλ 2πd ± nπ d± = λ 2 λ

and tan βl = tan

2πd 2πd ± nπ = tan λ λ

Some special cases are as follows: 1. ZL = 0 (i.e., a lossless line is short circuited) ⇒ Zin = j Z0 tan βl. 2. ZL = ∞ (i.e., a lossless line has an open circuit at the load) ⇒ Zin = −j Z0 cot βl. 3. l = λ/4, and therefore βl = π/2 ⇒ Zin = Z02 /ZL .

70

TRANSMISSION LINES

According to the first two cases, a lossless line can be used to synthesize an arbitrary reactance. The third case indicates that a quarter-wavelength-long line of suitable characteristic impedance can be used to transform a load impedance ZL to a new value of Zin . This type of transmission line, called an impedance transformer, is useful in impedance-matching applications. Further, this equation can be rearranged as follows: Z in =

1 ZL

= YL

Hence, normalized impedance at a point a quarter-wavelength away from the load is equal to the normalized load admittance. Example 3.5 A transmission line of length d and characteristic impedance Z0 acts as an impedance transformer to match a 150- load to a 300- line (see Figure 3.9). If the signal wavelength is 1 m, find (a) d, (b) Z0 , and (c) the reflection coefficient at the load. SOLUTION (a) d = λ/4 → d = 0.25 m √ (b) Z0 = Zin ZL → Z0 = (150 × 300)1/2 = 212.132 150 − 212.132 ZL − 1 ZL − Z0 = (c) L = = = 0.1716 ZL + Z0 150 + 212.132 ZL + 1 Example 3.6 Design a quarter-wavelength transformer to match a 20- load to the 45- line at 3 GHz. If this transformer is made from a Teflon-filled (εr = 2.1) coaxial line, calculate its length (in centimeters). Also, determine the diameter of its inner conductor if the inner diameter of the outer conductor is 0.5 cm. Assume that the impedance transformer is lossless. SOLUTION Z0 = and

√ Zin ZL = 45 × 20 = 30

Z0 =

L = C

b 200 × 10−9 ln = 30 −12 55.63 × εr × 10 a d

300 Ω Z0

Figure 3.9

Circuit arrangement for Example 3.5.

150 Ω

71

SENDING-END IMPEDANCE

Therefore, ln

b 30 = = 0.7251 a 41.3762

and 2b 0.5 2b b = = e0.7251 = 2.0649 ⇒ 2a = = = 0.2421 cm a 2a 2.0649 2.0649 √ Phase constant β = ω LC: 2π 1 ω√ LC = f × 200 × 55.63 × εr × 10−21 ⇒ = λ λ 2π √ 9 = 3 × 10 × 10−10 × 200 × 55.63 × 2.1 × 0.1

β=

= 14.5011 m−1 Therefore, λ = 0.06896 m and d = λ/4 = 0.01724 m = 1.724 cm. Example 3.7 Design a quarter-wavelength microstrip impedance transformer to match a patch antenna of 80 with a 50- line. The system is to be fabricated on a 1.6-mm-thick substrate (εr = 2.3) that operates at 2 GHz (see Figure 3.10). SOLUTION The characteristic impedance of the microstrip line impedance transformer must be Z0 = Z01 ZL = 63.2456 Design formulas for microstrip line are given in Appendix 2. Assume that the strip thickness t is less than 0.096 mm and that the dispersion is negligible for the time being at the operating frequency.

2.3 + 1 2

1/2

A=

63.2456 60

B=

60π2 = 6.1739 √ 63.2456 2.3

Z01

+

2.3 − 1 0.11 × 0.23 + = 1.4635 2.3 + 1 2.3

Z0

ZL

d

Figure 3.10 Circuit arrangement for Example 3.7.

72

TRANSMISSION LINES

Since A ≤ 1.52, 2 w = 6.1739 − 1 − ln(2 × 6.1739 − 1) h π 2.3 − 1 0.61 + ln(6.1739 − 1) + 0.39 − 2 × 2.3 2.3 =

2 × 3.2446 = 2.0656 ⇒ w = 3.3 mm π

At this point, we can check if the dispersion in the line is really negligible. For that, we determine the effective dielectric constant as follows: F

w h

−1/2 h −1/2 12 = 1 + 12 = 1+ = 0.383216 w 2.0656

Assuming that t/ h = 0.005, εe =

2.3 + 1 2.3 − 1 2.3 − 1 0.005 = 1.8981 ≈ 1.9 + × 0.383216 − ×√ 2 2 4.6 2.0656

and √

4h εr − 1 w 2 F = 0.5 + 1 + 2 × log 1 + = 0.213712 λ0 h √ 2 √ 2.3 − 1.9 √ + 1.9 = 1.909192 εe (f ) = 1 + 4 × F −1.5 Since εe (f ) is very close to εe , dispersion in the line can be neglected. Therefore, λ=

3 × 108 10.8821 = 2.72 cm √ m = 10.8821 cm ⇒ length of line = 9 4 2 × 10 × 1.9

Reflection Coefficient, Return Loss, and Insertion Loss The voltage reflection coefficient is defined as the ratio of reflected to incident phasor voltages at a location in the circuit. In the case of a transmission line terminated by load ZL , the voltage reflection coefficient is given by equation (3.2.4). Hence, =

Vref ZL − Z0 −2γl = e = ρL ej θ e−2(α+j β)l = ρL e−2αl e−j (2βl−θ) Vin ZL + Z0

(3.2.7)

73

SENDING-END IMPEDANCE 1.0

1.0

0.5 Imag. part

Imag. part

0.5

0.0

−0.5

−0.5 −1.0 −1.0

0.0

−0.5

0.0 Real part (a)

0.5

1.0

−1.0 −1.0

−0.5

0.0 Real part (b)

0.5

1.0

Figure 3.11 Reflection coefficient on (a) a lossy and (b) a lossless transmission line.

where ρL ej θ = (ZL − Z0 )/(ZL + Z0 ) is called the load reflection coefficient. Equation (3.2.7) indicates that the magnitude of reflection coefficient decreases by a factor of e−2αl as the observation point moves away from the load. Further, its phase angle changes by −2βl. A polar (magnitude and phase) plot of it will look like a spiral, as shown in Figure 3.11(a). However, the magnitude of reflection coefficient will not change if the line is lossless. Therefore, the reflection coefficient point will be moving clockwise on a circle of radius equal to its magnitude as the line length is increased. As illustrated in Figure 3.11 (b), it makes one complete revolution for each half-wavelength distance away from the load (because −2βλ/2 = −2π). Similarly, the current reflection coefficient, c , is defined as a ratio of reflected to incident signal-current phasors. It is related to the voltage reflection coefficient as follows: Iref −Vref /Z0 = = − c = Iin Vin /Z0 The return loss of a device is defined as the ratio of reflected power to incident power at its input. Since the power is proportional to the square of the voltage at that point, it may be found as return loss =

reflected power = ρ2 incident power

Generally, it is expressed in decibels as follows: return loss = 20 log10 ρ

dB

(3.2.8)

The insertion loss of a device is defined as the ratio of power transmitted (power available at the output port) to that of the power incident at its input. Since the power transmitted is equal to the difference of the incident and reflected

74

TRANSMISSION LINES

powers for a lossless device, the insertion loss can be expressed as follows: insertion lossof a lossless device = 10 log10 (1 − ρ2 )

dB

(3.2.9)

Low-Loss Transmission Lines Most practical transmission lines possess a very small loss of propagating signal. Therefore, expressions for the propagation constant and the characteristic impedance can be approximated for such lines as follows: γ=

√

ZY = (R + j ωL)(G + j ωC) =

−ω2 LC

R G 1+ 1+ j ωL j ωC

For R ωL and G ωC, a first-order approximation is √ γ = α + j β ≈ j ω LC 1 +

R G 1+ j 2ωL j 2ωC √ R G ≈ j ω LC 1 + + j 2ωL j 2ωC

Therefore, 1 C L R α≈ +G 2 L C and

√ β ≈ ω LC Z0 =

R + j ωL = G + j ωC

Np/m

rad/m

(3.2.10)

(3.2.11)

L G −1/2 R 1/2 1+ 1+ C j ωL j ωC

Hence,

L L R G R G Z0 ≈ 1+ 1− ≈ 1+ − C j 2ωL j 2ωC C j 2ωL j 2ωC L 1 R G 1+ − (3.2.12) = C j 2ω L C Thus, the attenuation constant of a low-loss line is independent of frequency, while its phase constant is linear, as in the case of a lossless line. However, the frequency dependency of its characteristic impedance is of concern to communication

75

SENDING-END IMPEDANCE

engineers, because it will distort the signal. If RC = GL, the frequency-dependent term will go to zero. This type of low-loss line is called a distortionless line. Hence, √ γ = ZY = (R + j ωL)(G + j ωC) RC C = α + jβ = (R + j ωL) ∵G = (3.2.13) L L Example 3.8 A signal propagating through a 50- distortionless transmission line attenuates at the rate of 0.01 dB/m. If this line has a capacitance of 100 pF/m, find (a) R, (b) L, (c) G, and (d) vp . SOLUTION Since the line is distortionless, L Z0 = = 50 C and

α = 0.01 dB/m ≈ 0.01/8.69 Np/m = 1.15 × 10−3 Np/m

Hence, (a) (b) (c) (d)

√ R = α L/C = 1.15 × 10−3 × 50 = 0.057 / m L = CZ02 = 10−10 × 502 H/m = 0.25 µH/m G = RC/L = R/Z02 = 0.057/502 S/m = 22.8µS/ m √ vp = 1/ LC = 2 × 108 m/s

Experimental Determination of Characteristic Impedance and Propagation Constant of a Transmission Line A transmission line of length d is kept open at one end, and the impedance at its other end is measured using an impedance bridge. Assume that it is Zoc . The process is repeated after placing a short circuit at its open end, and this impedance is recorded as Zsc . Using equation (3.2.5), one can write Zoc = Z0 coth γd

(3.2.14)

Zsc = Z0 tanh γd

(3.2.15)

and

Therefore, Z0 =

Zoc Zsc

and tanh γd =

(3.2.16)

1 Zsc ⇒ γ = tanh−1 Zoc d

Zsc Zoc

(3.2.17)

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These two equations can be solved to determine Z0 and γ. The following identity can be used to facilitate the evaluation of the propagation constant: tanh−1 Z =

1 1+Z ln 2 1−Z

The following examples illustrate this procedure. Example 3.9 Impedance at one end of the transmission line is measured to be Zin = 30 + j 60 using an impedance bridge, while its other end is terminated by a load ZL . The experiment is repeated twice, with the load replaced first by a short circuit and then by an open circuit. These data are recorded as j 53.1 and −j 48.3 , respectively. Find the characteristic resistance of this line and the load impedance. SOLUTION Z0 =

Zsc × Zoc = j 53.1 × (−j 48.3) = 50.6432

Zin = Z0 =

ZL + j Z0 tan βl ZL + Zsc = Z0 Z0 + j ZL tan βl Z0 + j ZL tan βl

Z0 ZL + Zsc j tan βl ZL + Z0 /(j tan βl)

Therefore, ZL + Zsc Zsc − Zin ⇒ ZL = Zoc ZL + Zoc Zin − Zoc j 53.1 − (30 + j 60) ZL = −j 48.3 × 30 + j 60 − (−j 48.3)

Zin = Zoc

Therefore, ZL = 11.6343 + j 6.3 Example 3.10 Measurements are made on a 1.5-m-long transmission line using an impedance bridge. After short-circuiting at one of its ends, impedance at the other end is found to be j 103 . Repeating the experiment with the short circuit now replaced by an open circuit gives −j 54.6 . Determine the propagation constant and the characteristic impedance of this line. SOLUTION Zoc Zsc = −j 54.6 × j 103 = 74.99 ≈ 75 j 103 tanh 1.5γ = = j 1.8969 ⇒ 1.5γ = tanh−1 j 1.8969 −j 54.3 Z0 =

77

SENDING-END IMPEDANCE

= 1.5γ =

1 1 + j 1.8969 ln 2 1 − j 1.8969 1 1 + j 1.8969 1 1 ln = ln(1 2.1713 rad) = ln(ej 2.1713 ) = j 1.08565 2 1 − j 1.8969 2 2

and γ = j 0.7238 m−1 . Example 3.11 A 10-m-long 50- lossless transmission line is terminated by a load, ZL = 100 + j 50 . It is driven by a signal generator that has an opencircuit voltage Vs at 100 0◦ V and source impedance Zs at 50 . The propagating signal has a phase velocity of 200 m/µS at 26 MHz. Determine the impedance at its input end and the phasor voltages at both its ends (see Figure 3.12). SOLUTION Zin = Z0 = 50

ZL + j Z0 tan βl Z0 + j ZL tan βl

and β =

ω 2π × 26 × 106 = = 0.8168 rad/m vp 200 × 106

100 + j 50 + j 50 tan(8.168) = 19.5278 0.181 rad 50 + j (100 + j 50) tan(8.168)

= 19.21 + j 3.52 The equivalent circuits at its input and at the load can be drawn as in Figure 3.13. Voltage at the input end, VA =

Vs 100 × 19.5278 0.181 rad Zin = Zs + Zin 50 + 19.21 + j 3.52

V

Zs

ZL

Vs

10 m

Figure 3.12

Zs

Vs

Circuit arrangement for Example 3.11. ZTh

VA

Zin

VTh

ZL

Figure 3.13 Equivalent circuits at the two ends of the transmission line shown in Figure 3.12.

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or 1952.78 0.181 rad ◦ = 28.1788 V 0.1302 rad = 28.1788 V 7.46 69.2995 0.0508 rad

VA =

For determining voltage at the load, a Th´evenin equivalent circuit can be used as follows: ZTh = Z0

Zs + j Z0 tan βl = Z0 Z0 + j Zs tan βl

(∵Zs = Z0 = 50 )

and VTh = (Vin + Vref )at

o.c.

= (2 × Vin )at

o.c.

= 100 V −8.168 rad

Therefore, VL =

VTh 100 −8.168 rad ZL = × (100 + j 50) V ZTh + ZL 50 + 100 + j 50

or ◦

◦

VL = 70.71 V −8.0261 rad = 70.71 V −459.86 = 70.71 V −99.86 Alternatively, V (z) = Vin e−j βz + Vref ej βz = Vin (e−j βz + ej βz )

where Vin is incident voltage at z = 0 and is the input reflection coefficient. ◦

Vin = 50 V 0 and =

19.21 + j 3.52 − 50 = 0.4472 2.9769 rad 19.21 + j 3.52 + 50

Hence, VL = V (z = 10 m) = 50(e−j 8.168 + 0.4472ej (2.9769+8.168) ) = 50 × 1.4142 V −1.743 rad or VL = 70.7117 V −99.8664◦ Example 3.12 Two identical signal generators are connected in parallel through a quarter-wavelength-long lossless 50- transmission line (Figure 3.14). Each of these generators has an open-circuit voltage of 12 0◦ V and a source impedance

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SENDING-END IMPEDANCE

Z0 = 50 Ω

Z0 = 50 Ω

50 Ω

A

50 Ω

12 V∠0°

B

12 V∠0°

λ/4

Figure 3.14

10 Ω

λ/4

Circuit arrangement for Example 3.12.

of 50 . It drives a 10- load through another quarter-wavelength-long similar transmission line. Determine the power dissipated by the load. SOLUTION The signal generator connected at the left (source A) and the quarter-wavelength-long line connecting it can be replaced by a Th´evenin equivalent voltage source of 12 V −90◦ with its internal resistance at 50 . This voltage source can be combined with the 12 0◦ V (source B) already there, which also has an internal resistance√of 50 . The resulting source will have a Th´evenin voltage of 6 − j 6 V = 6 2 V −45◦ and an internal resistance of 25 (i.e., two 50- resistances connected in parallel). The load will transform to 50 × 50/10 = 250 at the location of source B. Therefore, a simplified circuit can be drawn as in Figure 3.15. The voltage across 250 will be equal to 250 ×

√ 6 − j6 ◦ = 60 2/11 −45 V 275 V

Thus, the dissipated power, Pd , can be calculated as follows: Pd =

602 × 2 = 0.119008 W = 119.008 mW 112 × 2 × 250

25 Ω

6√2 V∠− 45°

Figure 3.15

250 Ω

Simplification of the circuit shown in Figure 3.14.

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TRANSMISSION LINES

3.3 STANDING WAVE AND STANDING WAVE RATIO Consider a lossless transmission line that is terminated by a load impedance ZL , as shown in Figure 3.16. Incident and reflected voltage phasors at its input (i.e., at z = 0) are assumed to be Vin and Vref , respectively. Therefore, total voltage V (z) can be expressed as follows: V (z) = Vin e−j βz + Vref e+j βz Alternatively, V (x) can be written as V (x) = V+ e+j βx + V− e−j βx = V+ (e+j βx + e−j βx ) or

V (x) = V+ [e+j βx + ρe−j (βx−φ) ]

where = ρej φ =

(3.3.1)

V− V+

V+ and V− represent incident and reflected wave voltage phasors, respectively, at the load point (i.e., at x = 0). Let us first consider two extreme conditions at the load. In one case, the transmission line has an open circuit, whereas in the other, it has a short-circuit termination. Therefore, the magnitude of the reflection coefficient, ρ, is unity in both cases. However, the phase angle, φ, is zero for the former case and π for the latter. Phasor diagrams for these two cases are depicted in Figure 3.17(a) and (b), respectively. As distance x from the load increases, the phasor ej βx rotates counterclockwise because of the increase in its phase angle βx. On the other hand, the phasor e−j βx rotates clockwise by the same amount. Therefore, the phase angle of the resulting voltage, V (x), remains constant with space coordinate x while its magnitude varies sinusoidally between ±2V+ . Since the phase angle of the resulting signal V (x) does not change with distance, it does not represent

V+ V−

Vin Vref

z

x

Figure 3.16 Lossless transmission line with termination.

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STANDING WAVE AND STANDING WAVE RATIO

e jβx βx V(x)

V(x) −βx

e−j(βx − π) π − βx

e−jβx

e jβx βx

(b)

(a)

Figure 3.17 Phasor diagrams of line voltage with (a) open-circuit and (b) short-circuit termination.

a propagating wave. Note that there are two waves propagating on this line in opposite directions. However, the resulting signal represents a standing wave. When the terminating load is of an arbitrary value that is different from the characteristic impedance of the line, there are still two waves propagating in opposite directions. The interference pattern of these two signals is stationary with time. Assuming that V+ is unity, a phasor diagram for this case is drawn as shown in Figure 3.18. The magnitude of the resulting signal, V (x), can be determined using the law of parallelograms as follows: |V (x)| = |V+ |[1 + ρ2 + 2ρ cos(2βx − φ)]1/2

ejβx βx

V(x)

θ

−(βx − φ) ρe−j(βx − φ)

(2βx − φ)

Figure 3.18 Phasor diagram of line voltage with arbitrary termination.

(3.3.2)

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The reflection coefficient, (x), can be expressed as (x) =

V− e−j βx = ρe−j (2βx−φ) V+ ej βx

(3.3.3)

Hence, the magnitude of this standing wave changes with location on the transmission line. Since x appears only in the argument of cosine function, the voltage magnitude has an extreme value whenever this argument is an integer multiple of π. It has a maximum whenever the reflected wave is in phase with the incident signal, which requires that the following condition be satisfied: 2βx − φ = ±2nπ,

n = 0, 1, 2, . . .

(3.3.4)

On the other hand, V (x) has a minimum value where the reflected and incident signals are out of phase. Hence, x satisfies the following condition at a minimum of the interference pattern: 2βx − φ = ±(2m + 1)π,

m = 0, 1, 2, . . .

(3.3.5)

Further, these extreme values of the standing waves are |V (x)|max = |V+ |(1 + ρ)

(3.3.6)

|V (x)|min = |V+ |(1 − ρ)

(3.3.7)

and

The ratio of maximum to minimum values of voltage V (x) is called the voltage standing wave ratio (VSWR). Therefore, VSWR = S =

|V (x)|max 1+ρ = |V (x)|min 1−ρ

(3.3.8)

Since 0 ≤ ρ ≤ 1 for a passive load, the minimum value of the VSWR will be unity (for a matched load) while its maximum value can be infinity (for total reflection, with a short circuit or an open circuit as the load). Assume that there is a voltage minimum at x1 from the load, and if one keeps moving toward the source, the next minimum occurs at x2 . In other words, there are two consecutive minimums at x1 and x2 with x2 > x1 . Hence, 2(βx1 − φ) = (2m1 + 1)π and 2(βx2 − φ) = [2(m1 + 1) + 1]π Subtracting the former equation from the latter, one finds that 2β(x2 − x1 ) = 2π ⇒ x2 − x1 =

λ 2

where x2 − x1 is the separation between the two consecutive minimums.

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STANDING WAVE AND STANDING WAVE RATIO

Similarly, it can be proved that two consecutive maximums are a half-wavelength apart and also that separation between the consecutive maximum and minimum is a quarter-wavelength. This information can be used to measure the wavelength of a propagating signal. In practice, the location of a minimum is preferred over that of a maximum. This is because minimums are sharper in comparison with maximums, as illustrated in Figure 3.19. Further, a short (or open) circuit must be used as the load for the best measurement accuracy. Measurement of Impedance Impedance of a one-port microwave device can be determined from measurement on the standing wave at its input. Those required parameters are the VSWR and the location of the first minimum (or maximum) from the load. A slotted line that is equipped with a detector probe is connected before the load to facilitate measurement. Since the output of a detector is proportional to power, the square root of the ratio is taken to find the VSWR. Since it may not be possible in most cases to probe up to the input terminals of the load, the location of the first minimum is determined as follows. An arbitrary minimum is located on the slotted line with unknown load. The load is then replaced by a short circuit. As a result, there is a shift in minimum, as shown in Figure 3.19. The shift of the original minimum away from the generator is equal to the location of the first minimum from the load. Since the reflected voltage is out of phase with that of the incident signal at the minimum of the standing wave pattern, a relation between the reflection coefficient, 1 = −ρ, and the impedance Z1 at this point may be written as 1 = −ρ =

2.0

Z1 − 1 Z1 + 1

With unknown load

With short circuit

Magnitude of V(x)

1.5

1.0

0.5

0.0 0.0

1.0

2.0

3.0

4.0

x

Figure 3.19

Standing wave pattern on a lossless transmission line.

5.0

84

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where Z 1 is the normalized impedance at the location of the first minimum of the voltage standing wave. Since S= and therefore, −

1+ρ 1−ρ

and ρ =

S−1 S+1

1 Z1 − 1 S−1 = ⇒ Z1 = S+1 S Z1 + 1

This normalized impedance is equal to the input impedance of a line that is terminated by load ZL and has a length of d1 . Hence, Z1 =

1 1 − j S tan βd1 Z L + j tan βd1 ⇒ ZL = = S S − j tan βd1 1 + j Z L tan βd1

Similarly, the reflected voltage is in phase with the incident signal at the maximum of the standing wave. Assume that the first maximum is located at a distance d2 from the unknown load and that the impedance at this point is Z2 . One can write S−1 Z2 − 1 2 = ρ = = ⇒ Z2 = S S+1 Z2 + 1 and ZL =

Z 2 − j tan βd2 1 − j Z 2 tan βd2

Since the maximum and minimum are measured on a lossless line that feeds the unknown load, magnitude of the reflection coefficient ρ does not change at different points on the line. Example 3.13 A load impedance of 73 − j 42.5 terminates a transmission line of characteristic impedance 50 + j 0.01 . Determine its reflection coefficient and the voltage standing wave ratio. SOLUTION = =

ZL − Z0 73 − j 42.5 − (50 + j 0.01) = ZL + Z0 73 − j 42.5 + (50 + j 0.01) 23 − j 42.51 48.3332 −1.0749 rad = 123 − j 42.49 130.1322 −0.3326 rad

◦

= 0.3714 −0.7423 rad = 0.3714 −42.53 and VSWR =

1+ρ 1 + 0.3714 = = 2.1817 1−ρ 1 − 0.3714

SMITH CHART

85

Note that this line is lossy, and therefore the reflection coefficient and VSWR will change with distance from the load. Example 3.14 A 100- transmission line is terminated by the load ZL . Measurements indicate that it has a VSWR of 2.4 and the standing wave minimums are 100 cm apart. The scale reading at one of these minimums is found to be 275 cm. When the load is replaced by a short circuit, the minimum moves away from the generator to a point where the scale shows 235 cm. Find the signal wavelength and the load impedance. SOLUTION Since the minimums are 100 cm apart, the signal wavelength λ = 2 × 100 = 200 cm = 2 m, β = π rad/m, and d1 = 275 − 235 = 40 cm. Hence, ZL =

1 − j 2.4 tan(0.4π) = 1.65 − j 0.9618 2.4 − j tan(0.4π)

or ZL = 165 − j 96.18 3.4 SMITH CHART Normalized impedance at a point in the circuit is related to its reflection coefficient as 1 + r + j i 1+ = Z = R + jX = (3.4.1) 1− 1 − r − j i where r and i represent real and imaginary parts of the reflection coefficient, while R and X are real and imaginary parts of the normalized impedance, respectively. This complex equation can be split into two, after equating real and imaginary components on its two sides. Hence, 2 2 1 R 2 + i = (3.4.2) r − 1+R 1+R and

2 1 2 1 = (r − 1) + i − X X 2

(3.4.3)

These two equations represent a family of circles on the complex -plane. The circle in (3.4.2) has its center at (R/(1 + R), 0) and a radius of 1/(1 + R). For R = 0 it is centered at the origin with unity radius. As R increases, the center of the constant-resistance circle moves on a positive real axis and its radius decreases. When R = ∞, the radius reduces to zero and the center of the circle moves to (1,0). These plots are shown in Figure 3.20. Note that for passive impedance, 0 ≤ R ≤ ∞ and −∞ ≤ X ≤ +∞. Similarly, (3.4.3) represents a circle that is centered at (1, 1/X) with a radius of 1/X. For X = 0, its center lies at (1,∞) with an infinite radius. Hence, it is a

86

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straight line along the r -axis. As X increases on the positive side (i.e., 0 ≤ X ≤ ∞), the center of the circle moves toward point (1,0) along a vertical line defined by r = 1, and its radius becomes smaller and smaller in size. For X = ∞, it becomes a point that is located at (1,0). Similar characteristics are observed for 0 ≥ X ≥ −∞. As shown in Figure 3.20, a graphical representation of these two equations for all possible normalized resistance and reactance values is known as the Smith chart. Thus, a normalized impedance point on the Smith chart represents the corresponding reflection coefficient in polar coordinates on the complex -plane. According to (3.2.7), the magnitude of the reflection coefficient on a lossless transmission line remains constant as ρL while its phase angle decreases as −2βl. Hence, it represents a circle of radius ρL . As one moves away from the load (i.e., toward the generator), the reflection coefficient point moves clockwise on this circle. Since the reflection coefficient repeats periodically at every half-wavelength, the circumference of the circle is equal to λ/2. For a given reflection coefficient, normalized impedance may be found using the impedance scale of the Smith chart. Further, R in (3.4.1) is equal to the VSWR for r > 0 and i = 0. On the other hand, it equals the inverse of VSWR for r < 0 and i = 0. Hence, R values on the positive r -axis also represent the VSWR.

Γi 90° (1,1)

Constant R or G circles

1.0 0.5

2.0

0.2

0°

180° 0

0.1

0.5

1.0

2.0

5.0

Constant X or B arcs 0.2 2.0 0.5 1.0 270° 1

Figure 3.20

Smith chart.

Γr

SMITH CHART

87

Since admittance represents the inverse of impedance, we can write Y =

1 Z

=

1− 1+

Hence, a similar analysis can be performed with the normalized admittance instead of impedance. It results in the same kind of chart except that the normalized conductance circles replace normalized resistance circles, while the normalized susceptance arcs replace normalized reactance. Normalized resistance (or conductance) of each circle is indicated on the r -axis of the Smith chart. Normalized positive reactance (or susceptance) arcs are shown on the upper half, while negative reactance (or susceptance) arcs are seen in the lower half. The Smith chart in conjunction with equation (3.2.7) facilitates the analysis and design of transmission line circuits. Example 3.15 A load impedance of 50 + j 100 terminates a lossless, quarterwavelength-long transmission line. If characteristic impedance of the line is 50 , find the impedance at its input end, the load reflection coefficient, and the VSWR on this transmission line. SOLUTION This problem can be solved using equations (3.2.6), (3.2.7), and (3.3.8) or the Smith chart. Let us try it both ways. 2π λ π ◦ × = = 90 λ 4 2 ZL + j Z0 tan βl (50 + j 100) + j 50 tan(90◦ ) Zin = Z0 = 50 Z0 + j ZL tan βl 50 + j (50 + j 100) tan(90◦ ) βl =

or Zin = 50

j 50 2500 = = 10 − j 20 j (50 + j 100) 50 + j 100

50 + j 100 − 50 j 100 100 90◦ ZL − Z0 = = = √ ZL + Z0 50 + j 100 + 50 100 + j 100 100 × 2 45◦ ◦ = 0.7071 45

=

VSWR =

1 + || 1 + 0.7071 1.7071 = = = 5.8283 1 − || 1 − 0.7071 0.2929

To solve this problem graphically using the Smith chart, normalized load impedance is determined as follows: ZL =

ZL 50 + j 100 = = 1 + j2 Z0 50

This point is located on the Smith chart as shown in Figure 3.21. A circle that passes through 1 + j 2 is then drawn with point 1 + j 0 as its center. Since the

88

1

0.9

0.7

0.15 0.35

2

4

0 0.2 30 0.

0.

THS TOWARD GENER E NG A VEL-- WAVELENGTHS TOWARD TOR A LO ---> - W AV D- W - WAV 0.0 0.0 5 0.0 -WA 0 is a perfect dielectric with εr = 2.25, and the region x < 0 is a free space. At the interface, subscript 1 denotes field components on the +x side of the boundary and subscript 2 those on the −x side. If D1 = 2 xˆ + 2yC/m ˆ , find D2 , E1 , and E2 . SOLUTION The conditions stated in the problem are illustrated in Figure 4.10. The interface is x = 0 plane, and therefore the x component, will be normal, with

Medium 2 Free space

y

Medium 1 εr = 2.25

x

z

Figure 4.10

Conditions for Example 4.8.

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ELECTROMAGNETIC FIELDS AND WAVES

the y component tangential to this plane. D1 = ε1 E1 → E1 =

D1 1 2 = xˆ + yˆ ε1 2.25ε0 2.25ε0

V/m

From the boundary conditions, the x component of electric flux density and y component of electric field intensity in medium 2 will be same as in medium 1. Hence, 2 D2x = 1 and E2y = 2.25ε0 Therefore, D2 = xˆ +

2 yˆ 2.25

C/m2

and E2 =

D2 1 2 = xˆ + yˆ ε0 ε0 2.25ε0

V/m

Example 4.9 A sphere of 2 m radius is made of a perfect dielectric material (medium 1). It is surrounded by free space (medium 2). Electric field intensities in the two media are given as follows: ˆ E1 = E01 (cos θˆr − sin θθ)

and E2 = E02

1+

8 r3

r ≤ 2m

4 cos θˆr − 1 − 3 sin θθˆ r

r ≥ 2m

Find the permittivity of the spherical medium. SOLUTION As indicated in Figure 4.11, the r components of the electric fields are normal and the θ components are tangential at the interface. Therefore, 4 sin θE02 → E02 = 2E01 − sin θE01 = − 1 − 8 z

θ r y

x

Figure 4.11

φ

Conditions for Example 4.9.

UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE

123

and ε1 E01 cos θ = ε0 E02 ž 2 cos θ → ε1 = 2ε0

E02 = 4ε0 E01

F/m

4.4 UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE Consider a uniform plane electromagnetic wave propagating in medium 1 as illustrated in Figure 4.12. The electric field Ei of this wave is y directed, and its magnetic field Hi is in the −x direction. A wave propagating in the +z direction is incident normally on the interface of medium 2 with medium 1. The electrical characteristics of the two media are as indicated in Figure 4.12. Since the wave experiences a change in electrical characteristics in its path, a part of the incident signal is reflected back, while the remaining signal is transmitted into medium 2. Thus, there are two waves in medium 1 propagating in opposite directions, and one in medium 2 that continues along +z. The electromagnetic fields associated with the incident wave can be expressed as follows: Ei = yE ˆ 0 e−j k1 z

(4.4.1)

and Hi = −xH ˆ 0 e−j k1 z = −xˆ

E0 −j k1 z e η1

(4.4.2)

x

Medium 1 σ1,ε1,µ1

Medium 2 σ2,ε2,µ2 Et

Ei Hr

Hi

Ht

Er z y

Figure 4.12 Geometry of a uniform plane wave incident normally on the interface of two media.

124

ELECTROMAGNETIC FIELDS AND WAVES

where E0 and H0 are arbitrary constants, k1 the complex wave number, and η1 the complex intrinsic impedance of medium 1. Similarly, electromagnetic fields associated with the reflected wave (Er and r H ) as well as the transmitted wave (Et and Ht ) can be expressed as follows: Er = yRE ˆ 0 e+j k1 z −j k1 z Hr = xRH ˆ = xR ˆ 0e

(4.4.3) E0 +j k1 z e η1

(4.4.4)

Et = yTE ˆ 0 e−j k2 z

(4.4.5)

and −j k1 z Ht = −xTH ˆ = −xT ˆ 0e

E0 −j k2 z e η2

(4.4.6)

where R and T are complex reflection and transmission coefficients, respectively, k2 the complex wave number in medium 2, and η2 the complex intrinsic impedance of medium 2. Now, applying the boundary conditions (tangential components of the fields are continuous across the boundary) at z = 0 gives E0 + RE 0 = TE 0 → 1 + R = T

(4.4.7)

and −

E0 E0 η1 E0 +R = −T →1−R = T η1 η1 η1 η2

(4.4.8)

Equations (4.4.7) and (4.4.8) can be solved for R and T as follows: T =

2η2 η1 + η2

(4.4.9)

R=

η2 − η1 η1 + η2

(4.4.10)

and

Therefore, the total electric field at any point in medium 1 is found to be ˆ 0 e−j k1 z + yRE ˆ 0 e+j k1 z E1 = yE |E1 |2 = |E0 |2 [e−2α1 z + ρ2 e2α1 z + 2ρ cos(2β1 z + θ)]

(4.4.11)

where R = ρej θ

(4.4.12)

k1 = β1 − j α1

(4.4.13)

UNIFORM PLANE WAVE INCIDENT NORMALLY ON AN INTERFACE

125

β1 is the phase constant in rad/s and α1 is the attenuation constant in Np/m for medium 1. It is to be noted that the propagation characteristics of this wave are very much similar to that of a wave propagating on the transmission line, as discussed in Chapter 3. Example 4.10 The electric field intensity of a uniform plane wave propagating in free space is given as Ei (z, t) = yˆ ž 4.8 cos(109 t − k0 z)

V/m

If this wave impinges normally on a lossless medium (µr = 1 and εr = 2.25) in the z ≥ 0 region, find k0 , k2 , R, T , the VSWR, and the average power transmitted into the medium. SOLUTION The wave numbers and intrinsic impedance of the wave in the two media are found as follows: √ k0 = ω µ0 ε0 = 109 4π × 10−7 × 8.854 × 10−12 = 3.3356 rad/m √ √ k2 = ω µ0 ε0 εr = k0 2.25 = 5.0034 rad/m

µ0 4π × 10−7 η1 = = = 376.7343 ε0 8.854 × 10−12 and η2 =

µ0 = ε0 εr

4π × 10−7 = 251.1562 8.854 × 10−12 × 2.25

Now R and T can be found via (4.4.9) and (4.4.10) as η2 − η1 ≈ −0.2 η1 + η2

R= and T =

2η2 ≈ 0.8 η1 + η2

The VSWR in medium 1 is found to be VSWR =

1+ρ 1 + 0.2 = = 1.5 1−ρ 1 − 0.2

The electric field intensity in medium 2 is Et (z, t) = yˆ ž 3.84 cos(109 t − k2 z)

V/m

126

ELECTROMAGNETIC FIELDS AND WAVES

Therefore, the time-averaged power transmitted into medium 2 may be found as follows: 1 1 |Et |2 1 3.842 = Pav = Re(E × H∗ ) = = 0.0294 W/m2 2 2 η2 2 251.1562 4.5 MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS Equation (4.1.9) implies that the magnetic flux does not start from or terminate at a magnetic charge the way that electric flux does with electric charge. This is because there is no real magnetic charge (or magnetic monopole). In fact, the smallest unit of the magnetic source is a magnetic dipole that is actually an infinitesimal current loop. Consider a general case where the source consists of two types of currents: a linear current that flows or oscillates in the linear direction, and an infinitesimal circulatory current that flows or oscillates around an infinitesimal loop. For this case it is rather hard to describe mathematically the electric current density J that includes both parts. Mathematically, it is sometimes advantageous to recognize its linear part as electric current Je and imagine its infinitesimal circulatory current as an equivalent magnetic dipole source. If the circulatory current varies in magnitude and direction or oscillates around the loop with time, the equivalent magnetic dipole strength also oscillates with time. When the magnetic dipole oscillates with time, it can be considered as an imaginary magnetic current flowing up and down. This is completely analogous to a time-varying electric dipole. Therefore, a time-varying circulatory current can be represented by an equivalent magnetic current and charges, Jm and ρm , respectively. Imagine now that the magnetic charge ρm produces a magnetic flux, analogous to the electric charge ρe producing the electric flux. Then equation (4.1.9) should be modified to ∇ ž B = ρm If (4.1.9) is modified, (4.1.6) also needs modification because divergence of the curl of a vector is always zero, which is not true because ρm is a function of time: ∂ ∂B ∂ ∇ ž (∇ × E) = −∇ ž = − (∇ ž B) = − ρm = 0 ∂t ∂t ∂t If it is assumed that the usual continuity equation should hold between ρm and Jm , ∇ ž Jm = −

∂ρm ∂t

Hence, equation (4.1.6) should be modified to ∇ × E = −J m −

∂B ∂t

MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS

127

Therefore, the modified Maxwell’s equations for the time-harmonic field and the two continuity equations can be expressed as: ∇ × E(r) = −Jm (r) − j ωB(r)

(4.5.1)

∇ × H(r) = Je (r) + j ω D(r)

(4.5.2)

∇ ž D(r) = ρe (r)

(4.5.3)

∇ ž B(r) = ρm (r)

(4.5.4)

∇ ž Je (r) = −j ωρe (r)

(4.5.5)

∇ ž Jm (r) = −j ωρm (r)

(4.5.6)

and

Magnetic Vector and Electric Scalar Potentials Potential functions are introduced to transform Maxwell’s equations mathematically and so facilitate solutions to electromagnetic problems. Consider that there are only electric sources (current and charge) present in a volume under consideration (i.e., Je = 0, ρe = 0, Jm = 0, and ρm = 0). Therefore, Maxwell’s equations for time-harmonic fields and the equation of continuity can be written as follows: ∇ × E = −j ωB

(4.5.7)

∇ × H = Je + j ωD

(4.5.8)

∇ ž D = ρe

(4.5.9)

∇žB = 0

(4.5.10)

∇ ž Je = −j ωρe

(4.5.11)

and

As defined earlier, the constitutive relations are B = µH

(4.5.12)

D = εE

(4.5.13)

and

Since divergence of the curl of a vector is always zero, the magnetic flux density in (4.5.10) may be assumed to be the curl of vector A. Hence, ∇žB = 0 ⇒ B = ∇ × A

(4.5.14)

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ELECTROMAGNETIC FIELDS AND WAVES

Substitution of B from (4.5.14) into (4.5.7), and recognizing the fact that the curl of the gradient of a scalar is always zero gives ∇ × E = −j ω ∇ × A ⇒ ∇×(E + j ωA) = 0 ⇒ E + j ωA = −∇φe Therefore, E = −∇φe − j ωA

(4.5.15)

where A is called the magnetic vector potential and φe is known as the electric scalar potential. Using (4.5.12), (4.5.13), and (4.5.15), (4.5.8) may be written as ∇×

∇×A = Je + j ωεE = Je + j ωε(−∇φe − j ωA) µ

If µ is not changing with space coordinates, this relation simplifies further as follows: ∇ × ∇ × A = ∇(∇ ž A) − ∇ 2 A = µJe − j ωεµ∇φe + ω2 µεA or ∇ 2 A + ω2 µεA = −µJe + j ωεµ∇φe + ∇(∇ ž A) or ∇ 2 A + ω2 µεA = −µJe + ∇(∇ ž A + j ωεµφe )

(4.5.16)

Similarly for ε not changing with space coordinates, (4.5.9), (4.5.13), and (4.5.15) give ∇ ž (−∇φe − j ωA) =

ρe ρe → ∇ 2 φe + j ω(∇ ž A) = − ε ε

(4.5.17)

Equations (4.5.16) and (4.5.17) represent a pair of coupled partial differential equations for φ and A. These may be uncoupled via Helmholtz’s theorem, which states that a vector is specified completely by its curl and divergence. Since only the curl of A is defined via (4.5.14), we are at liberty to specify its divergence. The Lorentz condition may be used to define it as ∇ ž A = −j ωεµφe

(4.5.18)

Thus, (4.5.16) and (4.5.17) simplify to ∇ 2 A + k 2 A = −µJ

(4.5.19)

ρe ε

(4.5.20)

and ∇ 2 φe + k 2 φe = −

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MODIFIED MAXWELL’S EQUATIONS AND POTENTIAL FUNCTIONS

where k 2 = ω2 µε and k is the wave number. It should be noted at this point that the continuity equation is implied by the Lorentz condition.

Electric Vector and Magnetic Scalar Potentials Electric vector and magnetic scalar potentials are introduced when the existence of only magnetic current and charge is assumed in the region (i.e, ρe = 0 and Je = 0). Therefore, Maxwell’s equations and the equation of continuity for this case are found to be ∇ × E = −Jm − j ωB

(4.5.21)

∇ × H = j ωD

(4.5.22)

∇žD = 0

(4.5.23)

∇ ž B = ρm

(4.5.24)

and ∇ ž Jm = −j ωρm

(4.5.25)

As before, D = εE and B = µH. Because of (4.5.23), it is assumed that D = −∇ × F

(4.5.26)

where F is called the electric vector potential. After substituting (4.5.26) into (4.5.22) and noting that curl of the gradient of a scalar is always zero, we get ∇ × H = −j ω∇ × F → ∇×(H + j ωF) = 0 → H + j ωF = −∇φm Therefore, H = −∇φm − j ωF

(4.5.27)

where φm is called the magnetic scalar potential. Now, from (4.5.21), (4.5.26), and (4.5.27),

∇×F ∇× − ε

= −Jm − j ωµ(−∇φm − j ωF)

or −∇ × ∇ × F = −εJm + j ωµε∇φm − ω2 µεF or −[∇(∇ ž F) − ∇ 2 F] + ω2 µεF = −εJm + j ωµε∇φm

(4.5.28)

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ELECTROMAGNETIC FIELDS AND WAVES

As in the preceding case, if the Lorentz condition ∇ ž F = −j ωµεφm

(4.5.29)

∇ 2 F + ω2 µεF = −εJm

(4.5.30)

is introduced, then Further, (4.5.24), (4.5.27), and (4.5.29) give ∇ ž [µ(−∇φm − j ωF)] = ρm ⇒ −∇ 2 φm − j ω(−j ωµεφm ) = or ∇ 2 φm + ω2 µεφm = −

ρm µ

ρm µ (4.5.31)

An electromagnetic field problem can be divided into these two cases, solved for each case, and the individual solutions can be superimposed to find the complete solution. Thus, the electromagnetic fields may be found easily after determining the vector potentials via (4.5.19) and (4.5.30) and the corresponding Lorentz conditions. This procedure is used in the following section to further formulate general solution techniques for source-free cases that are specialized subsequently for a few waveguides.

4.6 CONSTRUCTION OF SOLUTIONS The analysis presented in Section 4.5 is specialized here for a source-free region with rectangular as well as cylindrical coordinates. It is further divided into two cases; the electric vector potential is assumed to be zero in one, and the magnetic vector potential, zero in the other. Rectangular Coordinate System 1. Assume that A = zˆ Az and F = 0; then (4.5.14), (4.5.15), and (4.5.18) give

1 1 ∂Az ∂Az H = ∇×ˆzAz = xˆ + yˆ − + zˆ 0 (4.6.1) µ µ ∂y ∂x and E = −j ωˆzAz +

1 ∂Az ∇ j ωµε ∂z

(4.6.2)

For this case, (4.5.19) simplifies to ∇ 2 Az + k 2 Az = 0 →

∂ 2 Az ∂ 2 Az ∂ 2 Az + + + k 2 Az = 0 ∂x 2 ∂y 2 ∂z2

(4.6.3)

CONSTRUCTION OF SOLUTIONS

131

After (4.6.3) is solved for a given problem, field components are found from (4.6.1) and (4.6.2) as follows: Hx =

1 ∂Az µ ∂y

Hy = −

(4.6.4)

1 ∂Az µ ∂x

(4.6.5)

Hz = 0

(4.6.6) 2

Ex =

1 ∂ Az j ωµε ∂x∂z

(4.6.7)

Ey =

1 ∂ 2 Az j ωµε ∂y∂z

(4.6.8)

Ez =

1 j ωµε

and

∂2 2 + k Az ∂z2

(4.6.9)

Since the magnetic fields are transverse to the z-axis (Hz = 0), these fields are known as TMz (transverse magnetic to z ) mode fields. 2. Assume that F = zˆ Fz and A = 0; then (4.5.26) and (4.5.27) give

∂Fz 1 1 ∂Fz + yˆ − E = − ∇×ˆzFz = − xˆ + zˆ 0 ε ε ∂y ∂x

(4.6.10)

and H = −j ωˆzFz +

1 ∂Fz ∇ j ωµε ∂z

(4.6.11)

For this case, (4.5.30) simplifies to ∇ 2 Fz + k 2 Fz = 0 →

∂ 2 Fz ∂ 2 Fz ∂ 2 Fz + + + k 2 Fz = 0 ∂x 2 ∂y 2 ∂z2

(4.6.12)

After (4.6.12) is solved for a given problem, field components are found from (4.6.10) and (4.6.11) as follows: Ex = −

1 ∂Fz ε ∂y

(4.6.13)

1 ∂Fz ε ∂x Ez = 0

Ey =

Hx =

(4.6.14) (4.6.15) 2

1 ∂ Fz j ωµε ∂x ∂z

(4.6.16)

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ELECTROMAGNETIC FIELDS AND WAVES

Hy =

1 ∂ 2 Fz j ωµε ∂y∂z

Hz =

1 j ωµε

and

∂2 2 + k Fz ∂z2

(4.6.17)

(4.6.18)

Since the electric fields are transverse to the z-axis in this case, these fields are known as TE z (transverse electric to z ) mode fields. Cylindrical Coordinate System 1. Assume that A = zˆ Az and F = 0; then from (4.5.14), (4.5.15), and (4.5.18), we get

∂Az 1 1 1 ∂Az ˆ +φ − H = ∇×ˆzAz = ρˆ + zˆ 0 (4.6.19) µ µ ρ ∂φ ∂ρ and E = −j ωˆzAz +

∂Az 1 ∇ j ωµε ∂z

(4.6.20)

For this case, (4.5.19) in cylindrical coordinates may be written as follows: 1 ∂ ∂ 2 Az ∂Az 1 ∂ 2 Az 2 2 ∇ Az + k Az = 0 → + + k 2 Az = 0 (4.6.21) ρ + 2 ρ ∂ρ ∂ρ ρ ∂φ2 ∂zz Once solutions to (4.6.21) are found, electromagnetic fields may be determined from (4.6.19) and (4.6.20) as follows: Hρ =

1 ∂Az µρ ∂φ

(4.6.22)

1 ∂Az µ ∂ρ

(4.6.23)

Hφ = − Hz = 0

and

(4.6.24) 2

Eρ =

1 ∂ Az j ωµε ∂ρ∂z

(4.6.25)

Eφ =

1 ∂ 2 Az j ωµερ ∂φ∂z

(4.6.26)

1 Ez = j ωµε

∂2 2 + k Az ∂z2

(4.6.27)

METALLIC PARALLEL-PLATE WAVEGUIDE

133

This is the TMz mode (magnetic field transverse to the z-axis) in a cylindrical coordinate system. 2. For F = zˆ Fz and A = 0, (4.5.26) and (4.5.27) in cylindrical coordinates give

1 1 1 ∂Fz ∂Fz E = − ∇×ˆzFz = − ρˆ + φˆ − + zˆ 0 (4.6.28) ε ε ρ ∂φ ∂ρ and

1 ∂Fz H = −j ωˆzFz + ∇ j ωµε ∂z

(4.6.29)

Further, (4.5.30) reduces to 1 ∂ ∂ 2 Fz ∂Fz 1 ∂ 2 Fz + + k 2 Fz = 0 (4.6.30) ρ + 2 ∇ Fz + k Fz = 0 → ρ ∂ρ ∂ρ ρ ∂φ2 ∂z2 2

2

After (4.6.30) is solved, the field components are determined from (4.6.28) and (4.6.29) as follows: Eρ = −

1 ∂Fz ερ ∂φ

1 ∂Fz ε ∂ρ Ez = 0

Eφ =

(4.6.31) (4.6.32) (4.6.33)

2

Hρ =

1 ∂ Fz j ωµε ∂ρ∂z

(4.6.34)

Hφ =

1 ∂ 2 Fz j ωµερ ∂φ ∂z

(4.6.35)

Hz =

1 j ωµε

and

∂2 2 + k Fz ∂z2

(4.6.36)

This is the TEz mode (electric field transverse to the z-axis) in a cylindrical coordinate system. These formulations are used in the following sections to analyze propagation of electromagnetic signals through the waveguide. 4.7 METALLIC PARALLEL-PLATE WAVEGUIDE Consider a parallel-plate waveguide that consists of two perfectly conductive plates extending to infinity on the y –z plane, as illustrated in Figure 4.13. Separation between the two plates is assumed to be a. There is an electromagnetic

134

ELECTROMAGNETIC FIELDS AND WAVES z

a x

y

Figure 4.13 Metallic parallel-plate waveguide.

signal that propagates along the z-axis with a complex propagation constant γ. Therefore, the field variation along the z-axis is assumed to be e−γz , whereas it stays constant along the y-axis (i.e., ∂/∂y → 0). For analyzing TM modes between the two plates, partial differential equation (4.6.3) simplifies to the following ordinary differential equation: d 2 Az (x) + (k 2 + γ2 )Az (x) = 0 dx 2

(4.7.1)

A solution to this equation can be found easily as follows: Az (x) = c1 cos kx x + c2 sin kx x

(4.7.2)

where c1 and c2 are integration constants to be evaluated from the boundary conditions, and kx2 = k 2 + γ2 (4.7.3) Therefore,

Az (x, y, z) = (c1 cos kx x + c2 sin kx x)e−γz

(4.7.4)

Since the boundary conditions require that the tangential electric fields must be zero on the conducting surfaces at x = 0 and at x = a, Ez must be zero on these surfaces. It is to be noted that Ey is zero everywhere because Az is independent of y. For a nontrivial solution, Ez | x=0 = 0 → Az (x, y, z)| x=0 = 0 x=a

x=a

(4.7.5)

Equation (4.7.5) is satisfied only if c1 = 0

(4.7.6)

135

METALLIC PARALLEL-PLATE WAVEGUIDE

and kx =

mπ , a

Therefore, Az = c2 sin

m = 0, 1, 2, . . .

(4.7.7)

mπ x e−γz a

(4.7.8)

Further, the propagation constant can be found from (4.7.3) as follows: mπ 2 γ= − k2 (4.7.9) a Therefore, the signal will propagate without attenuation if k > mπ/a, and it will attenuate for k < mπ/a. The cutoff occurs at k = mπ/a. Once Az is determined, the field components may be found from (4.6.4)–(4.6.9) as follows: Hx = 0

(4.7.10)

mπ c2 mπ Hy = − cos x e−γz µ a a

(4.7.11)

Hz = 0

(4.7.12)

mπ γ mπ cos x e−γz Ex = −c2 j ωµε a a

(4.7.13)

Ey = 0

(4.7.14)

and Ez =

mπ c2 mπ 2 x e−γz sin j ωµε a a

(4.7.15)

These results are summarized in Table 4.1 assuming that −c2 (kc /µ) = E0 . A similar procedure may be used to determine the characteristics of TE mode fields propagating through this waveguide. Final results are summarized in Table 4.1. Example 4.11 A metallic parallel-plate waveguide is air-filled, and the separation between its plates is 4.5 cm. Investigate the characteristics of a 12-GHz signal propagating in TMm0 modes. SOLUTION kc =

mπ 2π 2a 2 × 4.5 × 10−2 = → λc = = a kc m m

fc =

3 × 108 3 × 108 × m = Hz = 3.3333m λc 9 × 10−2

m

and GHz

136

ELECTROMAGNETIC FIELDS AND WAVES

TABLE 4.1

Signal Propagation in Parallel-Plate Waveguides TEm0 Modes

TMm0 Modes

mπ a

mπ a

kc

γm0 Hz (x, z)

−

Hx (x, z)

kc H0 cos(kc x)e−γm0 z j ωµ 0

Ez (x, z) −

kc2 − k02

γm0 H0 sin(kc x)e−γm0 z j ωµ

kc2 − k02 0

j E0

kc sin(kc a)e−γm0 z ωε 0

Hy (x, z)

0

E0 cos(kc x)e−γm0 z

Ex (x, z)

0

γm0 E0 cos(kc x)e−γm0 z j ωε

Ey (x, z)

H0 sin(kc x)e−γm0 z

0

Hence, the cutoff frequencies for first four TM modes are found as follows: TM10 → fc = 3.3333 GHz TM20 → fc = 6.6667 GHz TM30 → fc = 10 GHz TM40 → fc = 13.3333 GHz Since the cutoff frequency for the TM40 mode is higher than the signal frequency of 12 GHz, only TM10 , TM20 , and TM30 modes will exist for this signal. The corresponding cutoff wavelengths are 9, 4.5, and 3 cm, respectively. The signal wavelength inside the guide for each mode can be found as follows: λ0 2.5 TM10 → λg = = cm = 2.6024 cm 2 1 − (λ0 /λc ) 1 − (2.5/9)2 λ0 2.5 TM20 → λg = = cm = 3.0067 cm 2 1 − (λ0 /λc ) 1 − (2.5/4.5)2 and λ0 2.5 TM30 → λg = = cm = 4.5227 cm 2 1 − (λ0 /λc ) 1 − (2.5/3)2

137

METALLIC RECTANGULAR WAVEGUIDE

4.8 METALLIC RECTANGULAR WAVEGUIDE In this section we present another application of the formulation described in Section 4.6 to analyze the electromagnetic signals propagating through a hollow metallic cylindrical waveguide of rectangular cross section, as shown in Figure 4.14. Both the geometry of a rectangular waveguide and the coordinate system are shown. It is assumed that its cross section is a × b, and the electromagnetic signal is propagating along the z-axis. Therefore, the field variation along the z-axis is given as e−j kz z . TM and TE modes propagating through this waveguide can be analyzed using equations (4.6.1) to (4.6.18). For TE modes, (4.6.12) reduces to ∂ 2 Fz (x, y) ∂ 2 Fz (x, y) + + (k 2 − kz2 )Fz (x, y) = 0 ∂x 2 ∂y 2

(4.8.1)

It is a second-order partial differential equation that can be solved using the separation-of-variables technique as follows. Assume that Fz (x, y) is a product of two functions, one dependent on x only and the other on y only. Hence, Fz (x, y) = f1 (x)f2 (y)

(4.8.2)

∂ 2 Fz (x, y) d 2 f1 (x) = f2 (y) 2 ∂x dx 2

(4.8.3)

d 2 f2 (y) ∂ 2 Fz (x, y) = f (x) 1 ∂y 2 dy 2

(4.8.4)

Therefore,

and

Substituting (4.8.2) to (4.8.4) into (4.8.1) and rearranging results in 1 d 2 f2 (y) 1 d 2 f1 (x) + = −(k 2 − kz2 ) f1 (x) dx 2 f2 (y) dy 2

(4.8.5)

y

b z

a x

Figure 4.14 Metallic rectangular waveguide.

138

ELECTROMAGNETIC FIELDS AND WAVES

In (4.8.5), the first term is only x dependent, the second is dependent on y only, and the sum of these two is always a constant, as its right-hand side indicates. Therefore, each term on the left-hand side must be independently constant as follows: 1 d 2 f1 (x) = −kx2 f1 (x) dx 2

(4.8.6)

1 d 2 f2 (y) = −ky2 f2 (y) dy 2

(4.8.7)

and

where kx and ky are constants. Equations (4.8.6) and (4.8.7) are in the form of the harmonic equation considered earlier, and its solutions are the harmonic functions. Complete solutions to each of these can be written as f1 (x) = c1 cos kx x + c2 sin kx x

(4.8.8)

f2 (y) = c3 cos ky y + c4 sin ky y

(4.8.9)

and

Therefore, the solution to (4.8.1) is found to be Fz (x, y) = (c1 cos kx x + c2 sin kx x)(c3 cos ky y + c4 sin ky y)

(4.8.10)

Further, equations (4.8.1), (4.8.6), and (4.8.7) give the separation equation, kx2 + ky2 + kz2 = k 2

(4.8.11)

Since tangential electric fields must be zero on the conducting surfaces, the boundary conditions for (4.8.1) are found to be ∂Fz y=0 Ex | = 0 → =0 (4.8.12) y=b ∂y y=0 y=b

and

∂Fz Ey x=0 = 0 → =0 x=a ∂x x=0

(4.8.13)

x=a

To satisfy these boundary conditions, (4.8.10) in conjunction with (4.8.12) and (4.8.14) gives c4 = 0 nπ ky = b

(4.8.14) (4.8.15)

METALLIC RECTANGULAR WAVEGUIDE

c2 = 0

139

(4.8.16)

and kx =

mπ a

(4.8.17)

where m and n are integers, including zero, except that both cannot be zero at the same time (which is a trivial case). Therefore, mπ nπ x cos y (4.8.18) Fz (x, y) = c1 c3 cos a b Further, kz may be found from (4.8.11) as kz =

k2 −

mπ 2 a

−

nπ 2 b

(4.8.19)

Consider a case when the waveguide is air-filled and therefore k = k0 is a pure real number. The following conditions are possible in this situation. (1) k02

mπ 2 a

+

nπ 2 b

(4.8.21)

In this case, the quantity under the square root in (4.8.19) is positive. Therefore, kz is real and the signal propagates through the waveguide without attenuation (ideally). This may be used to transmit the signal. (3) k02 =

mπ 2 a

+

nπ 2 b

= kc2

(4.8.22)

In this case, the quantity under the square root in (4.8.19) goes to zero. As a result, kz is zero (i.e., the cutoff condition). This wave number is known as the cutoff wave number kc , and the corresponding wavelength of the signal is called the cutoff wavelength λc . The corresponding frequency fc is called the cutoff frequency of the waveguide. Thus, the waveguide works as a high-pass filter. A signal propagates only when its frequency is greater than the cutoff frequency of the waveguide.

140

ELECTROMAGNETIC FIELDS AND WAVES

The wavelength and phase velocity of a signal propagating on the waveguide are found as follows:

and

2π 2π λ0 λg = = = 2 k0 1 − (kc /k0 ) 1 − (λ0 /λc )2 k02 − kc2 λ0 = 1 − (fc /f )2 vp =

ω f λ0 3 × 108 = = kz 1 − (λ0 /λc )2 1 − (λ0 /λc )2

m/s

(4.8.23)

(4.8.24)

where f is the signal frequency, k0 its wave number in free space, and λ0 the signal wavelength in free space. Fields inside the waveguide are found after substituting (4.8.18) into (4.6.13) to (4.6.18) as follows: Ex (x, y, z) = j H0 ωµky cos kx x sin ky y e−j kz z Ey (x, y, z) = −j H0 ωµkx sin kx x cos ky y e

−j kz z

Ez (x, y, z) = 0

and

(4.8.25) (4.8.26) (4.8.27)

mπ nπ Hx (x, y, z) = j H0 kz kx sin x cos y e−j kz z a b

(4.8.28)

Hy (x, y, z) = j H0 kz ky cos kx x sin ky ye−j kz z

(4.8.29)

Hz (x, y, z) = H0 kc2 cos kx x cos ky ye−j kz z

(4.8.30)

where H0 = −j c1 c3

1 ωµε

(4.8.31)

It is to be noted that if m and n are both zero, the fields do not exist. Therefore, for a nontrivial case, m or n has to be a nonzero integer. For a > b, the lowestorder mode that propagates is TE10 (i.e., for m = 1 and n = 0). It is known as the dominant mode for rectangular waveguides. From the expressions given in Table 4.2, it may be found easily that the lowest-order TM mode that can propagate through these waveguides is TM11 (i.e., for m = 1 and n = 1). For the TE10 mode, only the Hz , Hx , and Ey field components will be nonzero. Further, λc = 2a and λ0 λg(TE10 ) = (4.8.32) 1 − (λ0 /2a)2 Further,

π π sin x e−j kz z a a π j [(πx/a)+kz z] = −H0 ωµ0 − e−j [(πx/a)+kz z] e 2a

Ey (x, y, z) = −j H0 ωµ0

(4.8.33)

141

METALLIC RECTANGULAR WAVEGUIDE

TABLE 4.2 Section

Signal Propagation in Metallic Waveguides of Rectangular Cross

kc

TEmn Modes mπ 2 nπ 2 + a b

TMmn Modes mπ 2 nπ 2 + a b

k02 − kc2

k02 − kc2

λ0

λ0 1 − (kc /k0 )2

kz = −j γ λg

H0 kc2 cos

Hz (x, y)

Hy (x, y)

nπ mπ x cos y a b

0 nπ mπ x sin y a b mπ nπ nπ j E0 ωε sin x cos y b a b mπ nπ mπ −j E0 ωε cos x sin y a a b E0 kc2 sin

0

Ez (x, y) Hx (x, y)

1 − (kc /k0 )2

mπ nπ mπ sin x cos y a a b mπ nπ nπ cos x sin y j H0 kz b a b

j H0 kz

mπ nπ nπ cos x sin y b a b

−j

nπ mπ mπkz E0 cos x sin y a a b

mπ nπ mπ sin x cos y a a b

−j

nπ mπ nπkz E0 sin x cos y b a b

Ex (x, y)

j H0 ωµ0

Ey (x, y)

−j H0 ωµ0

On comparing this with (4.2.16), it may be found that this represents two plane electromagnetic waves propagating at angles ±θ after reflection from the sidewalls of the waveguide, as shown in Figure 4.15. This angle is found to be θ = sin−1

λ0 2a

(4.8.34)

Therefore, θ → 90◦ as λ0 → 2a (i.e., λc ), and the wave ceases to propagate.

x −θ a θ

z

Figure 4.15 Two uniform plane waves bouncing from the conducting sidewalls of a rectangular waveguide.

142

ELECTROMAGNETIC FIELDS AND WAVES

Example 4.12 The inside cross section of an air-filled rectangular metallic waveguide is 1.58 cm × 0.79 cm. (a) Determine its cutoff frequencies for TE10 , TE20 , TE01 , and TE11 modes. (b) Find the mode(s) that will propagate through this waveguide if the signal frequency is anywhere between 12 and 18 GHz. (c) If this waveguide is being used to send a 15.8-GHz signal in TE10 mode, find the phase velocity. SOLUTION (a) Since kc =

mπ 2 a

+

nπ 2 b

→ fc = 3 × 10

8

m 2 2a

+

n 2 2b

cutoff frequencies for various modes are found to be

2 2 0 1 + Hz = 9.4937 GHz TE10 → fc = 3 × 108 2 × 0.0158 2 × 0.0079

2 2 0 2 8 TE20 → fc = 3 × 10 + Hz = 18.9873 GHz 2 × 0.0158 2 × 0.0079

2 2 1 0 8 TE01 → fc = 3 × 10 + Hz = 18.9873 GHz 2 × 0.0158 2 × 0.0079 and

TE11 → fc = 3 × 108

1 2 × 0.0158

2 +

1 2 × 0.0079

2 Hz = 21.2285 GHz

(b) From part (a), the cutoff frequency for the TE10 mode is 9.4937 GHz. Since the next-higher modes have cutoff at 18.9873 GHz, only the TE10 mode will exist for the signal frequency band 12 to 18 GHz. 3 × 108 3 × 108 3 × 108 (c) vp = = = 1 − (λ0 /λc )2 1 − (fc /f )2 1 − (9.4937/15.8)2 = 3.7531 × 108 m/s 4.9 METALLIC CIRCULAR WAVEGUIDE Electromagnetic signal propagating through a hollow conducting cylindrical waveguide of circular cross section can be analyzed using (4.6.19) to (4.6.36). Figure 4.16 shows the geometry and coordinate systems for such analysis.

143

METALLIC CIRCULAR WAVEGUIDE x

φ a ρ z

y

Figure 4.16 Metallic circular waveguide geometry.

Assume that the waveguide radius is a and the field variation in the z-direction is e−j kz z (because the signal is propagating along z). For a TM mode, (4.6.21) gives ρ

∂ ∂Az (ρ, φ) ∂ 2 Az (ρ, φ) + (k 2 − kz2 )ρ2 Az (ρ, φ) = 0 ρ + ∂ρ ∂ρ ∂φ2

(4.9.1)

Following the technique used in Section 4.8, assume that Az (ρ, φ) is a product of two functions (one dependent on ρ only and the other on φ only) and solve (4.9.1) using the separation of variables. Hence, Az (ρ, φ) = f1 (ρ)f2 (φ)

(4.9.2)

Therefore, (4.9.1) may be written as 1 d 2 f2 (φ) ρ d df1 (ρ) = n2 ρ + (k 2 − kz2 )ρ2 = − f1 (ρ) dρ dρ f2 (φ) dφ2

(4.9.3)

where n is a constant that turns out to be an integer so that f2 (φ) is a single-valued harmonic function. After defining k 2 − kz2 = kρ2 for simplicity, (4.9.3) gives d df1 (ρ) ρ ρ + (kρ2 ρ2 − n2 )f1 (ρ) = 0 dρ dρ

(4.9.4)

d 2 f2 (φ) + n2 f2 (φ) = 0 dφ2

(4.9.5)

and

Equation (4.9.4) is known as Bessel’s equation of order n. Its solutions are the Bessel’s function of the first kind, the Neumann function (also known as the

144

ELECTROMAGNETIC FIELDS AND WAVES

Bessel’s function of the second kind), and the Hankel functions of the first and second kinds. Any two of these functions are linearly independent solutions to (4.9.4). Further, the first two of these represent standing waves, whereas the Hankel functions are convenient for traveling waves. For the present case, only Bessel’s functions of the first kind are possible solutions because Neumann functions go to infinity at the origin (i.e., ρ = 0). Equation (4.9.5) is the harmonic equation considered earlier. Therefore, Az (ρ, φ, z) = cJn (kρ ρ)

sin nφ cos nφ

(4.9.6)

Since tangential electric fields must be zero on the conducting wall at ρ = a, Jn (kρ a) = 0 → kρ =

pnm a

(4.9.7)

where pnm is the mth zero of Jn (kρ a).

TABLE 4.3

Signal Propagation in Circular-Cross-Section Waveguides

kc

TEnm Modes

TMnm Modes

pnm a

pnm a

k02 − kc2

kz = −j γmn

Hz (ρ, φ)

H0 Jn (kc ρ)

cos nφ

k02 − kc2 0

sin nφ

Ez (ρ, φ)

Hρ (ρ, φ)

Hφ (ρ, φ)

Eρ (ρ, φ)

Eφ (ρ, φ)

E0 Jn (kc ρ)

0

−j

−j

−j

kz H0 Jn (kc ρ) kc

nkz H0 Jn (kc ρ) ρkc2

sin nφ

nωµ H0 Jn (kc ρ) ρkc2

ωµ j H0 Jn (kc ρ) kc

cos nφ

− sin nφ

cos nφ − sin nφ

cos nφ cos nφ sin nφ

j

nωε E0 Jn (kc ρ) ρkc2

−j

−j

cos nφ sin nφ − sin nφ

ωε E0 Jn (kc ρ) kc kz E0 Jn (kc ρ) kc

nkz −j 2 E0 Jn (kc ρ) ρkc

cos nφ cos nφ sin nφ

cos nφ sin nφ

− sin nφ cos nφ

145

PROBLEMS

Similarly, (4.6.30) is solved to analyze TE modes in the circular waveguide. In this case, enforcement of the boundary condition (Eφ = 0 at ρ = a) gives ∂Jn (kρ ρ) = Jn (kρ a) = 0 (4.9.8) ∂(kρ ρ) ρ=a and kρ =

pnm a

(4.9.9)

is the mth root of (4.9.8). where pnm Characteristics of the TM and TE modes in a circular waveguide are listed in Table 4.3. Detailed derivations are deferred to Problems 4.34 and 4.35.

SUGGESTED READING Arfken, G., Mathematical Methods for Physicists. San Diego, CA: Academic Press, 1985. Balanis, C. A., Advanced Engineering Electromagnetics. New York: Wiley, 1989. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Harrington, R. F., Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill, 1961. Inan, U. S., and A. S. Inan, Electromagnetic Waves. Upper Saddle River, NJ: Prentice Hall, 2000.

PROBLEMS 4.1. The magnetic flux density on the y = 0 plane is given by B = 5 cos 3x cos 105 t yˆ

T

A square loop of conducting wire is placed on this plane with its vertices at (x, 0, 2), (x, 0, 3), (x + 1, 0, 3), and (x + 1, 0, 2). (a) Find the emf induced around the loop assuming that the loop is stationary. (b) If the loop is moving with a velocity of v = xˆ ž 104 m/s, determine the new induced emf. 4.2. If B = zˆ ž 5 cos(105 t − 0.5πx − πy)T, find the emf induced around a closed loop formed by connecting successively the points (0, 0, 0), (2, 0, 0), (2, 1, 0), (0, 1, 0), and (0, 0, 0). 4.3. For the following charge distribution, find the displacement flux that emanates from a cubical surface bounded by x = ±2, y = ±2, and z = ±2. ρ(x, y, z) = 5(3 − x 2 − y 2 − z2 )

C/m3

146

ELECTROMAGNETIC FIELDS AND WAVES

4.4. Find the charge densities that will produce the following electric flux densities: (a) D = xy xˆ + yzyˆ + zx zˆ C/m2 and (b) D = ρ sin φφˆ C/m2 . 4.5. In cylindrical coordinates, charge is distributed with uniform density 5 C/m3 within the region 1 < ρ < 2. Find the electric flux density in the region (a) 1 < ρ < 2 and (b) ρ > 2. 4.6. If the electric field intensity in free space is E = 5 cos(109 t − x + ky)ˆz V/m, find the value(s) of k. 4.7. The magnetic field intensity of an electromagnetic wave is given as H(x, t) = zˆ ž 0.6 cos βx cos 108 t

A/m

If the region is nonmagnetic with σ = 0 and ε = 6.25ε0 , find the corresponding electric field intensity and β. 4.8. A uniform plane electromagnetic wave is propagating in free space. Its instantaneous magnetic field intensity is given by π µA/m H = zˆ ž 5 cos 108 t − βy + 4 (a) Determine β and the signal wavelength. (b) Find the corresponding instantaneous electric field intensity. 4.9. A uniform plane wave of 50 MHz propagates in a lossless nonmagnetic medium in the +x direction. A probe located at x = 0 measures the phase angle of the wave to be 95◦ . An identical probe located at x = 2.5 m measures the phase to be −12◦ . What is the relative permittivity of the medium? 4.10. The magnetic field intensity of a uniform plane wave of 50 MHz is given as H(z) = (5xˆ + j 10y)e ˆ −j 2z A/m If it is in a lossless and nonmagnetic medium, find vp , η, and the instantaneous electric field intensity. 4.11. The electric field intensity of a uniform plane wave propagating in a nonmagnetic medium of zero conductivity is given as follows: E(y, t) = xˆ ž 12π cos(9π × 107 t + 0.3πy)

V/m

Find (a) the frequency, (b) the wavelength, (c) the direction of wave propagation, (d) the dielectric constant of the medium, and (e) the associated magnetic field intensity. 4.12. A uniform plane wave of 2 GHz is propagating in the +x direction in a nonmagnetic medium that has a dielectric constant of 2.25 and a loss tangent of 0.1. Its electric field is in the y-direction.

147

PROBLEMS

(a) Find the distance over which the amplitude of the propagating wave will reduce by 50 percent. (b) Find the intrinsic impedance, the wavelength, and the phase velocity of the wave in the medium. 4.13. A uniform plane wave propagates in the +z (downward) direction into the ocean (εr = 80, µr = 1, and σ = 4 S/m). Its magnetic field at the ocean surface (z = 0) is give as H(0, t) = yˆ ž 2 cos 109 t A/m (a) Determine the skin depth and intrinsic impedance of the ocean water. (b) Find the corresponding electric field intensity in the ocean.

4.14. A uniform plane wave of frequency 100 kHz is propagating in a material medium. It is given that (1) the fields attenuate by the factor e−1 over a distance of 1205 m, (2) the fields undergo a change in phase by 2π radians in a distance of 1321 m, and (3) the ratio of the amplitudes of its electric and magnetic field intensities at a point in the medium is 163.54 . Find the propagation constant γ and the intrinsic impedance η. 4.15. The electric field intensity of a uniform plane wave propagating through a lossless nonmagnetic medium is given by E(z, t) = xˆ ž 0.2 cos(109 t + πz)

V/m

Find (a) the direction of propagation, (b) the velocity of the signal, (c) the wavelength, and (d) the associated magnetic field intensity. 4.16. A material has an intrinsic impedance of 120 24◦ at 320 MHz. If its relative permeability is 4, find its (a) loss tangent, (b) dielectric constant, (c) conductivity, and (d) complex permittivity. 4.17. The electric field intensity of a uniform plane wave traveling through a material medium (εr = 4, σ = 0.1 S/m, µr = 1) is given by yˆ ž 10eγz V/m. If the signal frequency is 2.45 GHz, find the associated magnetic field intensity and propagation constant γ. 4.18. For a uniform plane wave propagating in the +z direction in a nonmagnetic material medium with σ = 10 S/m and εr = 9, the magnetic field intensity in the z = 0 plane is given by H(z = 0, t) = yˆ ž 0.1 cos(2π × 108 t)

A/m

Find the associated electric field intensity in the medium. 4.19. The magnetic field intensity of a uniform plane wave propagating through a certain nonmagnetic medium is given by H(x, t) = zˆ ž 5 cos(109 t − 6x)

A/m

148

ELECTROMAGNETIC FIELDS AND WAVES

Find (a) the direction of wave travel, (b) the velocity of wave, (c) the wavelength, and (d) the associated electric field intensity. 4.20. A uniform plane wave propagating through a lossless nonmagnetic medium has a power density of 8.05 W/m2 and an rms electric field intensity at 40.2 V/m. Find (a) the intrinsic impedance of the medium, (b) the rms magnetic field intensity, and (c) the velocity of the wave. 4.21. The electric field intensity of a signal transmitted through a coaxial line is given as follows: E = ρˆ

E0 −j βz e ρ

a≤ρ≤b

V/m

where a and b are inner and outer conductor radii, respectively, and β = ω(µε)0.5 . Find (a) the associated magnetic field intensity, (b) the surface current density Js , (c) the surface charge density ρs , and (d) the displacement current per unit length. 4.22. The region x > 0 is a perfect dielectric with εr = 6.25, while the region x < 0 is a perfect dielectric of εr = 2.25. At the interface, subscript 1 denotes field components on the +x side of the boundary and subscript 2 on the −x side. For D1 = 4xˆ + 2yˆ C/m2 , find D2 , E1 , and E2 . 4.23. A sphere of 1 m radius is made of perfect dielectric material (medium 1). It is surrounded by free space (medium 2). The electric field intensities in the two media are given as follows: ˆ E1 = E01 (cos θ rˆ − sin θ θ) and

E2 = E02

8 1+ 3 r

r ≤ 1m

4 cos θ rˆ − 1 − 3 r

sin θ θˆ

r ≥ 1m

Find the permittivity of the spherical medium. 4.24. A 150-MHz plane wave is normally incident from air (region 1) onto a semi-infinite nonmagnetic dielectric slab (region 2). If the voltage standing wave ratio in front of the slab is 2.5 and Emin is at the boundary, find (a) η2 , (b) εr2 , (c) the reflection coefficient, and (d) the distance d from the boundary to the nearest Emax of the standing wave pattern. 4.25. A uniform plane wave traveling through a nonmagnetic dielectric medium with εr = 4 is incident normally upon its interface with free space. If its electric field intensity is given by Ein = yˆ ž 3e−j 200πz

mV/m

149

PROBLEMS

Find (a) the corresponding magnetic field intensity, (b) the reflected electric field intensity, (c) the reflected magnetic field intensity, (d) the transmitted electric field intensity, and (e) the transmitted magnetic field intensity. 4.26. A uniform plane wave of 3 GHz propagating through free space is incident normally on a lossless dielectric medium with εr = 4 and µr = 1. The incident electric field at the interface has a value of 2 mV/m right before it strikes the interface. In free space, find (a) the reflection coefficient, (b) the VSWR, (c) the positions (in meters) of the maxima and minima of the electric field, and (d) the maximum and minimum values of the electric field. 4.27. The electric field intensity of a uniform plane wave propagating in air is Ei (z) = xˆ ž 5e−j 6z V/m. The wave is incident normally on an interface at z = 0 with a medium that has a dielectric constant at 6.25 and a loss tangent at 0.3. Find (a) the phasor expressions for reflected and transmitted electric and magnetic fields, and (b) the time-averaged power flow per unit area in the lossy medium. 4.28. Region 1 (x < 0) is air, whereas region 2 (x > 0) is a nonmagnetic medium characterized by σ = 10−4 S/m and εr = 4. A uniform plane wave with its electric field intensity as given below is incident on the interface at x = 0 from region 1.

2π x Ein = 5 cos 2π × 10 t − 3 8

V/m

Obtain the reflected and transmitted wave electric fields. 4.29. Verify the characteristics of the TE mode fields given in Table 4.1 for an electromagnetic signal that propagates in a metallic parallel-plate waveguide. 4.30. Separation between the plates of a parallel-plate waveguide is 0.625 cm, filled with a nonmagnetic dielectric material of εr = 4. Find the cutoff frequencies for the TM00 , TM10 , TM20 , TE10 , and TE20 modes. Can a signal propagate in the TE00 mode? Determine the phase velocities for all possible TE and TM modes that exist at 30 GHz. 4.31. Verify the characteristics of TM mode fields given in Table 4.2 for an electromagnetic signal that propagates in a metallic waveguide of rectangular cross section. 4.32. The wavelength of a signal propagating through an air-filled WR-340 rectangular waveguide is found to be 20 cm. Determine its frequency in gigahertz. 4.33. The inside cross section of an air-filled rectangular waveguide is 2.286 cm × 1.016 cm.

150

ELECTROMAGNETIC FIELDS AND WAVES

(a) Determine its cutoff frequencies for the TE10 , TE20 , TE01 , and TE11 modes. (b) Which mode(s) will propagate through this waveguide if the signal frequency is anywhere between 8 and 12 GHz? (c) Determine the phase velocity of a 9.375 GHz signal propagating in TE10 mode. 4.34. Verify the characteristics of TE mode fields given in Table 4.3 for an electromagnetic signal that propagates in a hollow metallic cylindrical waveguide of circular cross section. 4.35. Verify the characteristics of the TM mode fields given in Table 4.3 for an electromagnetic signal that propagates in a hollow metallic cylindrical waveguide of circular cross section.

5 RESONANT CIRCUITS

A communication circuit designer frequently requires means to select (or reject) a band of frequencies from a wide signal spectrum. Resonant circuits provide such filtering. There are well-developed, sophisticated methodologies to meet virtually any specification. However, a simple circuit suffices in many cases. Further, resonant circuits are an integral part of the frequency-selective amplifier as well as of the oscillator designs. These networks are also used for impedance transformation and matching. In this chapter we describe the analysis and design of these simple frequencyselective circuits and present the characteristic behavior of series and parallel resonant circuits. Related parameters, such as quality factor, bandwidth, and input impedance, are introduced that will be used in several subsequent chapters. Transmission lines with an open or short circuit at their ends are considered next, and their relationships with the resonant circuits are established. Transformer-coupled parallel resonant circuits are discussed briefly because of their significance in the radio-frequency range. In the final section we summarize the design procedure for rectangular and circular cylindrical cavities, and the dielectric resonator.

5.1 SERIES RESONANT CIRCUITS Consider the series R –L–C circuit shown in Figure 5.1. Since the inductive reactance is directly proportional to signal frequency, it tries to block the highfrequency contents of the signal. On the other hand, capacitive reactance is Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

151

152

RESONANT CIRCUITS L

C

vin(t)

R

vo(t)

Figure 5.1 Series R –L–C circuit with input–output terminals.

inversely proportional to the frequency. Therefore, it tries to stop its lower frequencies. Note that the voltage across an ideal inductor leads the current by 90◦ (i.e., the phase angle of an inductive reactance is 90◦ ). In the case of a capacitor, voltage across its terminals lags behind the current by 90◦ (i.e., the phase angle of a capacitive reactance is −90◦ ). This means it is possible that the inductive reactance will be canceled out by the capacitive reactance at some intermediate frequency. This frequency is called the resonant frequency of the circuit. If the input signal frequency is equal to the resonant frequency, maximum current will flow through the resistor and it will be in phase with the input voltage. In this case, the output voltage Vo will be equal to the input voltage Vin . It can be analyzed as follows: From Kirchhoff’s voltage law, t L dvo (t) 1 vo (t) dt + vo (t) = vin (t) (5.1.1) + R dt RC −∞ Taking the Laplace transform of this equation with initial conditions as zero (i.e., no energy storage initially), we get sL 1 (5.1.2) + + 1 Vo (s) = Vin (s) R sRC where s is the complex frequency (Laplace variable). The transfer function of this circuit, T (s), is given by T (s) =

Vo (s) 1 sR = = 2 Vin (s) sL/R + 1/sRC + 1 s L + sR + 1/C

(5.1.3)

Therefore, the transfer function of this circuit has a zero at the origin of the complex s-plane and also has two poles. The location of these poles can be determined by solving the quadratic equation s 2 L + sR +

1 =0 C

Two possible solutions to this equation are as follows: 2 R 1 R − ± s1,2 = − 2L 2L LC

(5.1.4)

(5.1.5)

SERIES RESONANT CIRCUITS

153

The circuit response will be influenced by the location of these poles. Therefore, these networks can be characterized as follows: √ √ 1. If R/2L > 1/ LC (i.e., R > 2 L/C), both of these poles will be real and distinct, and the circuit is overdamped. √ √ 2. If R/2L = 1/ LC (i.e., R = 2 L/C), the transfer function will have √ double poles at s = −R/2L = −1/ LC. The circuit is critically damped. √ √ 3. If R/2L < 1/ LC (i.e., R < 2 L/C), the two poles of T (s) will be complex conjugate of each other. The circuit is underdamped. Alternatively, the transfer function may be rearranged as follows: T (s) =

sCRω20 sCR = s 2 LC + sRC + 1 s 2 + 2ζω0 s + ω20

(5.1.6)

where R ζ= 2 ω0 = √

C L

1 LC

(5.1.7) (5.1.8)

ζ is called the damping ratio and ω0 is the undamped natural frequency. Poles of T (s) are determined by solving the equation s 2 + 2ζω0 s + ω20 = 0

(5.1.9)

For ζ < 1, s1,2 = −ζω0 ± j ω0 1 − ζ2 . As shown in Figure 5.2, the two poles are complex conjugates of each other. The output transient response will be oscillatory with a ringing frequency of ω0 (1 − ζ2 ) and an exponentially decaying amplitude. This circuit is underdamped. For ζ = 0, the two poles move on the imaginary axis. Transient response will be oscillatory. It is a critically damped case. For ζ = 1, the poles are on the negative real axis. Transient response decays exponentially. In this case, the circuit is overdamped. Consider the unit step function shown in Figure 5.3. It is like a direct voltage source of 1 V that is turned on at time t = 0. If it represents input voltage vin (t), the corresponding output vo (t) can be determined via Laplace transform technique. The Laplace transform of a unit step at the origin is equal to 1/s. Hence, the output voltage, vo (t), is found as follows: vo (t) = L−1 Vo (s) = L−1

sCRω20 CRω20 1 −1 = L s 2 + 2ζω0 s + ω20 s (s + ζω0 )2 + (1 − ζ2 )ω20

154

RESONANT CIRCUITS jw

w0(1 − ζ2)1/2

w0

−σ −w0(1 − ζ2)1/2

−ζw0

Figure 5.2 Pole–zero plot of the transfer function.

vin(t)

1V

Time

t→∞

t=0

Figure 5.3

Unit-step input voltage.

where L−1 represents the inverse Laplace transform operator. Therefore, 2ζ vo (t) = e−ζω0 t sin(ω0 t 1 − ζ2 )u(t) 1 − ζ2 This response is illustrated in Figure 5.4 for three different damping factors. As can be seen, initial ringing lasts longer for a lower damping factor. A sinusoidal steady-state response of the circuit can be determined easily after replacing s by j ω, as follows: Vo (j ω) =

Vin (j ω) Vin (j ω) = j ωL/R + 1/j ωRC + 1 (1 + j/RC)(LCω − 1/ω)

Vo (j ω) =

Vin (j ω) Vin (j ω) = 2 (1 + j/ω RC)(ω/ω (1 + j/RC)(ω/ω0 − 1/ω) 0 0 − ω0 /ω)

or

155

SERIES RESONANT CIRCUITS 0.8

ζ = 0.95

0.6

Output voltage

0.4 ζ = 0.5

0.2

ζ = 0.1

0.0

−0.2

10

0

20 w0t

30

40

50

Figure 5.4 Response of a series R –L–C circuit to a unit step input for three different damping factors.

The quality factor, Q, of the resonant circuit is a measure of its frequency selectivity. It is defined as Q = ω0

average stored energy power loss

(5.1.10)

Hence, 1

Q = ω0 12 2

LI 2 I 2R

=

ω0 L R

Since ω0 L = 1/ω0 C, √ ω0 L LC 1 1 L 1 Q= = = = = R ω0 RC RC R C 2ζ Therefore, Vo (j ω) =

Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω)

(5.1.11)

(5.1.12)

Alternatively, 1 Vo (j ω) = A(j ω) = Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω)

(5.1.13)

The magnitude and phase angle of (5.1.13) are illustrated in Figures 5.5 and 5.6, respectively. Figure 5.5 shows that the output voltage is equal to the input for

156

RESONANT CIRCUITS

1.0

Q=1

0.8

Magnitude

0.6

Q = 10

0.4 Q = 100 0.2 ω0 0.0 0.0

20.0

40.0

60.0

80.0

100.0

ω

Figure 5.5

Magnitude of A(j ω) as a function of ω.

2.0 Q = 100

1.5

Phase angle (rad)

1.0 0.5 Q = 10

0.0

Q=1

−0.5 −1.0 −1.5 −2.0 0.0

ω0 20.0

40.0

60.0

80.0

100.0

ω

Figure 5.6 Phase angle of A(j ω) as a function of ω.

a signal frequency equal to the resonant frequency of the circuit. Further, phase angles of the two signals in Figure 5.6 are the same at this frequency, irrespective of the quality factor of the circuit. As signal frequency moves away from this point on either side, the output voltage decreases. The rate of decrease depends on the quality factor of the circuit. For higher Q, the magnitude is sharper,

157

SERIES RESONANT CIRCUITS

indicating a higher selectivity of the circuit. If signal frequency is below the resonant frequency, the output voltage leads the input. For a signal frequency far below the resonance, the output leads the input by almost 90◦ . On the other hand, it lags behind the input for higher frequencies. It converges to −90◦ as the signal frequency moves far beyond the resonant frequency. Thus, the phase angle changes between π/2 and −π/2, following a sharper change around the resonance for high-Q circuits. Note that the voltage across the series-connected inductor and capacitor combined has inverse characteristics to those of the voltage across the resistor. Mathematically, VLC (j ω) = Vin (j ω) − Vo (j ω) where VLC (j ω) is the voltage across the inductor and capacitor combined. In this case, sinusoidal steady-state response can be obtained as follows: VLC (j ω) Vo (j ω) 1 =1− =1− Vin (j ω) Vin (j ω) 1 + j Q(ω/ω0 − ω0 /ω) =

j Q(ω/ω0 − ω0 /ω) 1 + j Q(ω/ω0 − ω0 /ω)

Hence, this configuration of the circuit represents a band-rejection filter. Half-power frequencies ω1 and ω2 of a bandpass circuit can be determined from (5.1.13) as 1 1 = ⇒ 2 = 1 + Q2 2 2 1 + Q (ω/ω0 − ω0 /ω)2 Therefore,

ω0 ω − Q ω0 ω

= ±1

Assuming that ω1 < ω0 < ω2 ,

ω1 ω0 Q − ω0 ω1

= −1

and

ω0 ω2 − Q ω0 ω2 Therefore,

=1

ω0 ω1 ω0 ω2 − =− − ω0 ω2 ω0 ω1

ω0 ω − ω0 ω

2

158

RESONANT CIRCUITS

or ω2 ω2 ω2 ω2 ω2 − 0 = −ω1 + 0 ⇒ (ω2 + ω1 ) = 0 + 0 = ω20 ω2 ω1 ω1 ω1

1 1 + ω1 ω2

or ω20 = ω1 ω2

(5.1.14)

and ω2 ω0 1 ω0 ω1 − = − ⇒ ω1 − 0 = − ω0 ω1 Q ω1 Q or ω1 − ω2 = −

ω0 ω0 ⇒Q= Q ω2 − ω1

(5.1.15)

Example 5.1 Determine the element values of a resonant circuit that passes all the sinusoidal signals from 9 to 11 MHz. This circuit is to be connected between a voltage source with negligible internal impedance and a communication system with its input impedance at 50 . Plot its characteristics in a frequency band of 1 to 20 MHz. SOLUTION From (5.1.14), ω0 =

√

ω1 ω2 → f0 =

√ f1 f2 = 9 × 11 = 9.949874 MHz

From (5.1.11) and (5.1.15), Q=

ω0 R ω0 L = ⇒L= R ω1 − ω2 ω1 − ω2 50 = 3.978874 × 10−6 H ≈ 4 µH = 2π × 106 × (11 − 9)

From (5.1.8), ω0 = √

1 LC

⇒C=

1 = 6.430503 × 10−11 F ≈ 64.3 pF Lω20

The circuit arrangement is shown in Figure 5.7. Its magnitude and phase characteristics are displayed in Figure 5.8. Input Impedance Impedance across the input terminals of a series R –L–C circuit can be determined as follows: ω20 1 Zin = R + j ωL + (5.1.16) = R + j ωL 1 − 2 j ωC ω

159

SERIES RESONANT CIRCUITS L = 4 µH

C = 64.3 pF

R = 50 Ω

v

Figure 5.7

Filter circuit arrangement for Example 5.1.

1.0

Magnitude of A(jω)

0.8

0.6

0.4

0.2

0.0

0

5

10 ω (a)

15

0

5

10 ω (b)

15

20

2.0

Phase angle in (rad)

1.5 1.0 0.5 0.0 −0.5 −1.0 −1.5 −2.0

20

Figure 5.8 Magnitude (a) and phase (b) plots of A(j ω) for the circuit in Figure 5.7.

160

RESONANT CIRCUITS

At resonance, the inductive reactance cancels out the capacitive reactance. Therefore, the input impedance reduces to the total resistance of the circuit. If signal frequency changes from the resonant frequency by ±δω, the input impedance can be approximated as Zin = R + j ωL

(ω + ω0 )(ω − ω0 ) 2QRδω ≈ R + j 2δωL = R + j (5.1.17) 2 ω ω0

Alternatively, Zin ≈ R + j 2δωL =

ω0 L ω0 + j 2(ω − ω0 )L = j 2 ω − ω0 + L Q j 2Q

1 L = j 2 ω − ω0 1 + j 2Q

(5.1.18)

Therefore, a series resonant circuit can be analyzed with R as zero (i.e., assuming that the circuit is lossless). The losses can be included subsequently by replacing a real resonant frequency, ω0 , by the complex frequency, ω0 [1 + j (1/2Q)]. At resonance, the current through the circuit, Ir , is Ir =

Vin R

(5.1.19)

Therefore, voltages across the inductor, VL , and the capacitor, VC , are VL = j ω0 L

Vin = j QVin R

(5.1.20)

and VC =

1 Vin = −j QVin j ω0 C R

(5.1.21)

Hence, the magnitude of voltage across the inductor is equal to the quality factor times input voltage, while its phase leads by 90◦ . The magnitude of the voltage across the capacitor is the same as that across the inductor. However, it is 180◦ out of phase because it lags behind the input voltage by 90◦ .

5.2 PARALLEL RESONANT CIRCUITS Consider an R –L–C circuit in which the three components are connected in parallel, as shown in Figure 5.9. A subscript p is used to differentiate the circuit elements from those used in series circuit of Section 5.1. A current source, iin (t), is connected across its terminals and io (t) is current through the resistor Rp .

161

PARALLEL RESONANT CIRCUITS io(t) iin(t)

Lp

Figure 5.9

Cp

Rp

vo(t)

Parallel R –L–C circuit.

Voltage across this circuit is vo (t). From Kirchhoff’s current law, iin (t) =

1 Lp

t −∞

Rp io (t) dt + Cp

d(Rp io (t)) + io (t) dt

(5.2.1)

Assuming that no energy was stored in the circuit initially, we take the Laplace transform of (5.2.1). It gives Iin (s) = Hence,

Rp + sRp Cp + 1 Io (s) sLp

Io (s) sLp /Rp = 2 Iin (s) s Lp Cp + s(Lp /Rp ) + 1

(5.2.2)

Note that this equation is similar to (5.1.6). It changes to T (s) if RC replaces Lp /Rp . Therefore, the results of the series resonant circuit can be used for this parallel resonant circuit, provided that ζ=

1 2 ω0 Rp Cp

and 1 ω0 = Lp Cp Hence, 1 1 = ζ= 2 ω0 Rp Cp 2Rp

(5.2.3) Lp Cp

(5.2.4)

The quality factor, Qp , and the impedance, Zp , of the parallel resonant circuit can be determined as follows: Qseries = Zp =

ω0 L Rp 1 = ⇒ Qp = ω0 Rp Cp = R ω0 RC ω0 Lp

(5.2.5)

I0 (j ω)Rp j ωLp Vo (j ω) = = (5.2.6) 2 Iin (iω) Iin (j ω) −ω Lp Cp + j ω(Lp /Rp ) + 1

162

RESONANT CIRCUITS

Input Admittance Admittance across input terminals of the parallel resonant circuit (i.e., the admittance seen by the current source) can be determined as follows: Yin =

ω2 1 1 1 1 = + j ωCp + = + j ωCp 1 − 02 Zp Rp j ωLp Rp ω

(5.2.7)

Hence, input admittance will be equal to 1/Rp at the resonance. It will become zero (i.e., the impedance will be infinite) for a lossless circuit. It can be approximated around the resonance, ω0 ± δω, as Yin ≈

1 1 2δωQ + j 2δωCp = +j Rp Rp ω0 Rp

(5.2.8)

The corresponding impedance is Zp ≈

Rp 1 + j (2Qδω/ω0 )

(5.2.9)

Current through the capacitor, Ic , at the resonance is Ic = j ω0 Cp Rp Iin = j QIin

(5.2.10)

and current through the inductor, IL , is IL =

1 Rp Iin = −j QIin j ω0 Lp

(5.2.11)

Thus, current through the inductor is equal in magnitude but opposite in phase to that through the capacitor. Further, these currents are larger than the input current by a factor of Q. Quality Factor of a Resonant Circuit If resistance R represents losses in the resonant circuit, Q given by the preceding formulas is known as the unloaded Q. If the power loss due to external load coupling is included through an additional resistance RL , the external Qe is defined as follows: ω L 0 RL Qe = R L ω0 Lp

for series resonant circuit (5.2.12) for parallel resonant circuit

163

PARALLEL RESONANT CIRCUITS

The loaded Q, QL , of a resonant circuit includes internal losses as well as the power extracted by the external load. It is defined as follows: ω0 L RL + R for series resonant circuit

QL = (5.2.13)

Rp R L for parallel resonant circuit ω0 Lp where

RL

Rp =

RL Rp RL + Rp

Hence, the following relation holds good for both types of resonant circuit: 1 1 1 = + QL Qe Q

(5.2.14)

See Table 5.1. Example 5.2 Consider the loaded parallel resonant circuit in Figure 5.10. Compute the resonant frequency in radians per second, unloaded Q, and the loaded Q of this circuit. SOLUTION

TABLE 5.1

1 1 ω0 = =√ = 108 rad/s Lp Cp 10−5 × 10−11 Relations for Series and Parallel Resonant Circuits Series

Parallel

1

LC R C 2 L

1 Lp Cp Lp 1 2Rp Cp

Unloaded Q

1 ω0 L = R ω0 RC

Rp = ω0 Rp Cp ω0 Lp

External Q = Qe

1 ω0 L = RL ω0 RL C

RL = ω0 RL Cp ω0 Lp

Loaded Q = QL

QQe Q + Qe

QQe Q + Qe

ω0 Damping factor, ζ

Input impedance, Zin , around resonance

√

R+j

2RQδω ω0

R 1 + j (2Qδω/ω0 )

164

RESONANT CIRCUITS

Load 10 pF

10 µH

100 kΩ

100 kΩ

Figure 5.10

The unloaded

Circuit for Example 5.2.

Q=

105 Rp = 8 = 100 ω0 Lp 10 × 10−5

Qe =

RL 105 = 8 = 100 ω0 Lp 10 × 10−5

The external Q,

The loaded Q,

Rp

RL QQe 50 × 103 QL = = = 8 = 50 ω0 Lp Q + Qe 10 × 10−5

5.3 TRANSFORMER-COUPLED CIRCUITS Transformers are used as a means of coupling as well as of impedance transforming in electronic circuits. Transformers with tuned circuits in one or both of their sides are employed in voltage amplifiers and oscillators operating at radio frequencies. In this section we present an equivalent model and an analytical procedure for the transformer-coupled circuits. Consider a load impedance ZL that is coupled to the voltage source Vs via a transformer as illustrated in Figure 5.11. Source impedance is assumed to be Zs . The transformer has a turns ratio of n : 1 between its primary (the source) and secondary (the load) sides. Using the notations indicated, equations for various voltages and currents can be written in phasor form as follows: V1 = j ωL1 I1 + j ωMI2

(5.3.1)

V2 = j ωMI1 + j ωL2 I2

(5.3.2)

where M is the mutual inductance between the two sides of the transformer. Standard convention with a dot on each side is used. Hence, magnetic fluxes reinforce each other for the case of currents entering this terminal on both sides, and M is positive.

165

TRANSFORMER-COUPLED CIRCUITS I1

I2

+

+

Zs V1 Vs

L1

ZL

V2

L2 n:1

−

Figure 5.11

−

Transformer-coupled circuit.

The following relations hold for an ideal transformer operating at any frequency: V1 = nV2 I2 I1 = − n and

(5.3.3) (5.3.4)

nV2 V2 V1 = n2 = Z1 = = n2 Z2 I1 −I2 /n −I2

(5.3.5)

Several equivalent circuits are available for a transformer. We consider one of these that is most useful in analyzing the communication circuits. This equivalent circuit is illustrated in Figure 5.12. The following equations for phasor voltages and currents may be formulated using the notations indicated in the figure. I2 I2 V1 = j ω(1 − ξ)L1 I1 + j ωξL1 I1 + (5.3.6) = j ωL1 I1 + j ωξL1 n n I1 + Zs

I2/n

I2 +

(1 − ξ) L1 n:1

V1

V2

ξ L1

ZL

Vs −

− Ideal transformer

Figure 5.12 Equivalent model of the transformer-coupled circuit shown in Figure 5.11.

166

RESONANT CIRCUITS

and

1 I2 V2 = j ωξL1 I1 + n n

(5.3.7)

If the circuit shown in Figure 5.12 is equivalent to that shown in Figure 5.11, these two equations represent the same voltages as those of (5.3.1) and (5.3.2). Hence, ξL1 =M (5.3.8) n and

ξL1 = L2 n2

In other words,

n=

and ξ=

(5.3.9)

ξL1 L2

(5.3.10)

√ M ξ M nM =√ ⇒ ξ= √ =κ L1 L1 L2 L1 L2

(5.3.11)

where κ is called the coefficient of coupling. It is close to unity for a tightly coupled transformer, and close to zero for a loose coupling. Example 5.3 A tightly coupled transformer is used in the circuit shown in Figure 5.13. Inductances of its primary and secondary sides are 320 and 20 nH, respectively. Find its equivalent circuit, the resonant frequency, and the Q of this circuit.

I1 +

6 pF Is

V1

−

Figure 5.13

I2 +

L1

V2

30.7 pF

L2

−

Transformer-coupled resonant circuit.

62.5 kΩ

167

TRANSFORMER-COUPLED CIRCUITS

+

Is

+

V1

R1

C1

V2

L1 n:1

−

− Ideal transformer

Figure 5.14 Equivalent model of the transformer-coupled circuit shown in Figure 5.13.

SOLUTION Since the transformer is tightly coupled, κ ≈ 1. Therefore, ξ ≈ 1, 1 − ξ ≈ 0, and its equivalent circuit simplifies as shown in Figure 5.14. From (5.3.10), ξL1 320 n= =4 ≈ L2 20 and from (5.3.5), Y2 1 Z1 = n Z2 ⇒ Y1 = 2 = 2 n n 2

1 + j ωC2 R2

Therefore, R1 = n2 R2 = 16 × 62.5 k = 1 M 1 C1 = 6 + 2 30.7 pF = 6 + 1.91875 ≈ 7.92 pF 4 1 1 =√ = 628.1486 × 106 rad/s ω0 = √ −9 L1 C1 320 × 10 × 7.92 × 10−12 ω0 = 99.97 MHz ≈ 100 MHz f0 = 2π and Q=

106 R1 = = 4974.9375 ω0 L1 628.1486 × 106 × 320 × 10−9

Example 5.4 A transformer-coupled circuit is shown in Figure 5.15. Draw its equivalent circuit using an ideal transformer. (a) If the transformer is tuned only to its secondary side (i.e., R1 and C1 are removed), determine its resonant frequency and impedance seen by the current source.

168

RESONANT CIRCUITS I1

I2

+

R1

C1

Is

V1

+

C2

V2

L1

L2

R2

−

−

Figure 5.15

I0

Double-tuned transformer-coupled circuit.

(b) Determine the current transfer function (I0 /Is ) for the entire circuit (i.e., R1 and C1 are included in the circuit). If the transformer is loosely coupled, and both of the sides have identical quality factors as well as the resonant frequencies, determine the locations of the poles of the current transfer function on the complex ω-plane. SOLUTION Following the preceding analysis and Figure 5.12, the equivalent circuit can be drawn easily, as shown in Figure 5.16. Using notations as indicated in the figure, circuit voltages and currents can be found as follows: I2 nV2 = sξL1 I1 + (5.3.12) n R2 = R2 I0 1 + sR2 C2 1 + sC1 [nV2 + s(1 − ξ)L1 I1 ] Is = I1 + R1

V2 = −I2

I1 +

Is

C1

−

ξL1

(5.3.14)

I2

+

(1 − ξ)L1 R1

V1

I2/n

(5.3.13)

+

I0

n:1 nV2

V2

−

C2

−

Ideal transformer

Figure 5.16 Equivalent representation for the circuit shown in Figure 5.15.

R2

169

TRANSFORMER-COUPLED CIRCUITS

From (5.3.12), (5.3.13), and (5.3.10), we find that sL2 sξL1 s 2 ξL1 R2 C2 + sξL1 2 I1 = 1 + s L C + V R2 I0 = 1+ 2 2 2 2 n R2 R2 n or

sξL1 /nR2 I0 = 2 I1 1 + s L2 C2 + sL2 /R2

(5.3.15)

When R1 and C1 are absent, I1 will be equal to Is , and (5.3.15) will represent the current transfer characteristics. This equation is similar to (5.2.2). Hence, the resonant frequency is found as follows: ω0 = √

1 L2 C2

(5.3.16)

Note that the resonant frequency in this case depends only on L2 and C2 . It is independent of inductance L1 of the primary side. The impedance seen by the current source, with R1 and C1 removed, is determined as follows: Zin (s) =

V1 n2 sL2 = sL1 (1 − ξ) + 2 I1 s L2 C2 + s(L2 /R2 ) + 1

(5.3.17)

Zin (j ω0 ) = j ω0 L1 (1 − ξ) + n2 R2

(5.3.18)

At resonance, and if transformer is tightly coupled, ξ ≈ 1, Zin (j ω0 ) = n2 R2

(5.3.19)

When R1 and C1 are included in the circuit, (5.3.12) to (5.3.14) can be solved to obtain a relationship between I0 and Is . The desired current transfer function can be obtained as I0 sξL1 /nR2 = 2 Is [s L2 C2 + s(L2 /R2 ) + 1]{1 + s(1 − ξ)L1 [(1/R1 ) + sC1 ]} +sξL1 [(1/R1 ) + sC1 ]

(5.3.20)

Note that the denominator of this equation is now a polynomial of the fourth degree. Hence, this current ratio has four poles on the complex ω-plane. If the transformer is loosely coupled, ξ will be negligible. In that case, the denominator of (5.3.20) can be approximated as follows: I0 sξL1 /nR2 ≈ 2 Is [s L2 C2 + s(L2 /R2 ) + 1]{1 + sL1 [(1/R1 ) + sC1 ]} + sξL1 [(1/R1 ) + sC1 ] (5.3.21)

170

RESONANT CIRCUITS

For ω1 = ω2 = ω0 , and Q1 = Q2 = Q0 , L1 C1 = L2 C2 =

1 ω20

and R1 R2 = = ω0 Q0 L1 L2 Hence, (5.3.21) may be written as sξL1 /nR2 I0 ≈ 2 2 Is (s /ω0 + s/ω0 Q + 1)2 + ξ(s 2 /ω20 + s/ω0 Q)

(5.3.22)

The poles of (5.3.22) are determined by solving the equation

2 2 s2 s s s + + 1 + ξ + =0 ω0 Q0 ω0 Q0 ω20 ω20

(5.3.23)

The roots of this equation are found to be 1 2 ω0 ± j ω0 (1 ∓ ξ) − s≈− 2Q0 2Q0

(5.3.24)

5.4 TRANSMISSION LINE RESONANT CIRCUITS Transmission lines with open- or short-circuited ends are frequently used as resonant circuits in the ultrahigh frequency and microwave range. We consider such networks in this section. Since the quality factor is an important parameter of these circuits, we need to include the finite (even though small) loss in the line. There are four basic types of these networks, as illustrated in Figure 5.17. It can be found easily by analyzing the input impedance characteristics around the resonant wavelength, λr , that the circuits of Figure 5.17 (a) and (d) behave like a series R –L–C circuit. On the other hand, the other two transmission lines possess the characteristics of a parallel resonant circuit. A quantitative analysis of these circuits is presented below for n as unity. Short-Circuited Line Consider a transmission line of length l and characteristic impedance Z0 . It has a propagation constant, γ = α + j β. The line is short circuited at one of its ends, as shown in Figure 5.18. The impedance at its other end, Zin , can be determined

171

TRANSMISSION LINE RESONANT CIRCUITS (2n − 1)λr 4

l=

l=

γ = α + jβ Open Z0

Z0

Open γ = α + jβ

(a) l=

n λr 2

(b)

(2n − 1)λr 4

l=

γ = α + jβ

n λr 2

Z0

Short

Short γ = α + jβ

Z0 (c)

(d )

Figure 5.17 Four basic types of transmission line resonant circuits, n = 1, 2, . . . .

l

Zin →

Z0

γ = α + jβ

Figure 5.18 Short-circuited lossy transmission line.

from (3.2.5) as follows: Zin = Z0 tanh γl = Z0 tanh(α + j β)l = Z0

tanh αl + j tan βl 1 + j tanh αl tan βl

(5.4.1)

For αl 1, tanh αl ≈ αl, and assuming that the line supports only the TEM mode, we find that βl = ωl/vp . Hence, it can be simplified around the resonant frequency, ω0 , as follows: βl =

ωl ω0 l δωl δωl = + = βr l + vp vp vp vp

(5.4.2)

where βr is the phase constant at the resonance. If the transmission line is one-half wavelength long at the resonant frequency, βr l = π and

l π = . vp ω0

172

RESONANT CIRCUITS

Therefore, tan βl can be approximated as follows: δω l π δω π δω π δω tan βl = tan π + ≈ = tan π + = tan vp ω0 ω0 ω0 and Zin ≈ Z0

αl + j (π δω/ω0 ) π δω ≈ Z0 αl + j 1 + j (αl)(π δω/ω0 ) ω0

(5.4.3)

It is assumed in (5.4.3) that (αl)(π δω/ω0 ) 1. For a series resonant circuit, Zin ≈ R + j 2δωL. Hence, a half-wavelength-long transmission line with short-circuit termination is similar to a series resonant circuit. The equivalent-circuit parameters are found as follows (assuming that losses in the line are small such that the characteristic impedance is a real number): R ≈ Z0 αl = 12 Z0 αλr

(5.4.4)

π Z0 2 ω0 2 C≈ πω0 Z0 L≈

(5.4.5) (5.4.6)

and Q=

π βr ω0 L ≈ = R 2αl 2α

(5.4.7)

On the other hand, if the transmission line is a quarter-wavelength long at the resonant frequency, βr l = π/2, and l/vp = π/2 ω0 . Therefore,

π δωl tan βl = tan + 2 vp and Zin ≈ Z0

π πδω = tan + 2 2 ω0

πδω = − cot 2 ω0

Z0 /αl −j (2 ω0 /πδω) = 1 − j αl(2 ω0 /πδω) 1 + j (π/αl)(δω/2 ω0 )

≈−

2 ω0 πδω (5.4.8) (5.4.9)

This expression is similar to the one obtained for a parallel resonant circuit in (5.2.9). Therefore, this transmission line is working as a parallel resonant circuit with equivalent parameters as follows: Rp ≈

Z0 4Z0 = αl αλr 4Z0 Lp ≈ πω0 π Cp ≈ 4 ω0 Z0

(5.4.10) (5.4.11) (5.4.12)

TRANSMISSION LINE RESONANT CIRCUITS

173

and Q=

βr π = 4αl 2α

(5.4.13)

Open-Circuited Line The analysis of the open-circuited transmission line shown in Figure 5.19 can be performed following a similar procedure. These results are summarized below. From (3.2.5), Z0 Zin |ZL =∞ = (5.4.14) tanh γl For a transmission line that is a half-wavelength long, the input impedance can be approximated as follows: Zin ≈ Z0

1 + j (αl)(πδω/ω0 ) Z0 ≈ αl + j (πδω/ω0 ) αl + j (πδω/ω0 )

(5.4.15)

This is similar to (5.2.9). Therefore, a half-wavelength-long open-circuited transmission line is similar to a parallel resonant circuit with equivalent parameters as follows: Z0 2Z0 = αl αλr 2Z0 Lp ≈ πω0 π Cp ≈ 2 ω0 Z0

Rp ≈

(5.4.16) (5.4.17) (5.4.18)

and Q=

βr π = 2αl 2α

(5.4.19)

If the line is only a quarter-wavelength long, the input impedance can be approximated as 1 − j (αl)(2 ω0 /πδω) πδω Zin ≈ Z0 (5.4.20) ≈ Z0 αl + j αl − j (2 ω0 /πδω) 2 ω0 l

Zin →

Z0

γ = α + jβ

Figure 5.19 Open-circuited transmission line.

174 TABLE 5.2

RESONANT CIRCUITS

Equivalent-Circuit Parameters of the Resonant Lines for n = 1 Quarter-wavelength line

Resonance R L C Q

Half-wavelength line

Open Circuit

Short Circuit

Open Circuit

Short Circuit

Series

Parallel

Parallel

Series

4Z0 αλr 4Z0 πω0 π 4Z0 ω0 βr 2α

2Z0 αλr 2Z0 πω0 π 2Z0 ω0 βr 2α

1 Z αλr 4 0

πZ0 4 ω0 4 πZ0 ω0 βr 2α

1 Z αλr 2 0

πZ0 2 ω0 2 πZ0 ω0 βr 2α

Since (5.4.20) is similar to (5.1.17), this transmission line works as a series resonant circuit. Its equivalent-circuit parameters are found as follows (see Table 5.2): R ≈ Z0 αl = 14 Z0 αλr πZ0 4 ω0 4 C≈ πω0 Z0 L≈

(5.4.21) (5.4.22) (5.4.23)

and Q=

π β0 πZ0 = = 4R 4αl 2α

(5.4.24)

Example 5.5 Design a half-wavelength-long coaxial line resonator that is shortcircuited at its ends. Its inner conductor radius is 0.455 mm and the inner radius of the outer conductor is 1.499 mm. The conductors are made of copper. Compare the Q value of an air-filled resonator to that of a Teflon-filled resonator operating at 5 GHz. The dielectric constant of Teflon is 2.08 and its loss tangent is 0.0004. SOLUTION From the relations for coaxial lines given in Appendix 2, ωµ0 2π × 5 × 109 × 4π × 10−7 Rs = = = 0.018427 2σ 2 × 5.813 × 107 Rs 1 1 αc = + √ b 2 ln(b/a) µ0 /ε0 a

175

TRANSMISSION LINE RESONANT CIRCUITS

0.018427 = 2 × 376.7343 × ln(1.499/0.455)

100 100 + 0.455 1.499

= 0.058768 Np/m With Teflon filling, Rs αc = √ 2 µ0 /ε0 εr ln(b/a)

1 1 + a b

√ 100 100 0.018427 × 2.08 + = 2 × 376.7343 × ln(1.499/0.455) 0.455 1.499

= 0.084757 Np/m √ ω√ µ0 ε0 εr tan δ = π × 5 × 109 × 3 × 108 × 2.08 × 0.0004 αd = 2 = 0.030206 Np/m 2π × 5 × 109 = 104.719755 rad/m 3 × 108 √ 2π × 5 × 109 × 2.08 = 151.029 rad/m βd = 3 × 108 β0 104.719755 Qair = = = 890.96 2α 2 × 0.058768 βd 151.029 QTeflon = = = 656.8629 2α 2 × (0.084757 + 0.030206) β0 =

Example 5.6 Design a half-wavelength-long microstrip resonator using a 50- line that is short circuited at its ends. The substrate thickness is 0.159 cm, with its dielectric constant 2.08 and the loss tangent 0.0004. The conductors are made of copper. Compute the length of the line for resonance at 2 GHz and the Q of the resonator. Assume that thickness t of the microstrip is 0.159 µm. SOLUTION Design equations for the microstrip line (from Appendix 2) are 8eA e2A − 2 εr − 1 w B − 1 − ln(2B − 1) + = 2 2εr h 0.61 π × ln(B − 1) + 0.39 − εr where Z0 A= 60

εr + 1 2

1/2

for A > 1.52

for A ≤ 1.52

εr − 1 0.11 + 0.23 + εr + 1 εr

176

RESONANT CIRCUITS

w

h

Figure 5.20 Geometry of the microstrip line.

and B=

60π2 √ Z0 εr

See Figure 5.20. Therefore, 50 A= 60

2.08 + 1 2

1/2

2.08 − 1 0.11 + 0.23 + = 1.1333 2.08 + 1 2.08

Since A is smaller than 1.52, we need B to determine the width of microstrip. Hence, B=

60π2 = 8.212 √ 50 × 2.08

w 2 2.08 − 1 = 8.212 − 1 − ln(2 × 8.212 − 1) + h π 2 × 2.08 0.61 × ln(8.212 − 1) + 0.39 − = 3.1921 2.08 Therefore,

w = 0.507543 cm ≈ 0.51 cm

Further, εre = 0.5 2.08 + 1 + (2.08 − 1) × √ −

1 1 + 12/3.192094

(2.08 − 1)0.0001 √ = 1.7875 4.6 3.1921

Therefore, the length of the resonator can be determined as follows: l=

vp /f c 3 × 1010 λ = = √ cm = 5.6096 cm √ = 2 2 2f εe 2 × 2 × 109 × 1.7875

177

MICROWAVE RESONATORS

Since σcopper = 5.813 × 107 S/m, the quality factor of this resonator is determined as follows. For w/ h ≥ 1/2π, w t h we = + 0.3979 1 + ln 2 = 3.1923 h h h t and

1.25t 1.25 h h 1− + ln 2 = 2.5476 ζ=1+ we πh π t

Therefore, for w/ h ≥ 1, αc is found to be fGHz we 0.667 (we / h) −5 ζZ0 εre αc = 44.1255 × 10 + h σ h we / h + 1.444

Np/m

= 0.0428 Np/m Similarly, αd is found from (A2.28) as follows: αd = 10.4766

εr εre − 1 fGHz tan δ √ εr − 1 εre

Np/m = 0.0095 Np/m

and β=

2π 2π 2π = = rad/m = 56.0035 rad/m λ 2l 2 × 5.6096/100

Hence, Q of the resonator is found as Q=

56.0035 β = = 535.4 2α 2 × (0.0428 + 0.0095)

5.5 MICROWAVE RESONATORS Cavities and dielectric resonators are commonly employed as resonant circuits at frequencies above 1 GHz. These resonators provide much higher Q than the lumped elements and transmission line circuits. However, the characterization of these devices requires analysis of associated electromagnetic fields. Characteristic relations of selected microwave resonators are summarized in this section. Interested readers can find several excellent references and textbooks analyzing these and other resonators. Rectangular Cavities Consider a rectangular cavity made of conducting walls with dimensions a × b × c, as shown in Figure 5.21. It is filled with a dielectric material of dielectric constant εr and relative permeability µr . In general, it can support both TEmnp

178

RESONANT CIRCUITS

y c

b

x a

z

Figure 5.21 Geometry of the rectangular cavity resonator.

and TMmnp modes, which may be degenerate. The cutoff frequency of TEmn and TMmn modes can be determined from the formula n 2 m 2 300 fc (MHz) = √ + MHz µr εr 2a 2b The resonant frequency fr of a rectangular cavity operating in either TEmnp or TMmnp mode can be determined as follows: 300 fr (MHz) = √ µr εr

m 2 2a

n 2 p 2 + + 2b 2c

MHz

(5.5.1)

If the cavity is made of a perfect conductor and filled with a perfect dielectric, it will have infinite Q. However, it is not possible in practice. In the case of cavity walls having a finite conductivity (instead of infinite for the perfect conductor) but filled with a perfect dielectric (no dielectric loss), its quality factor Qc for TE10p mode is given by Qc |10p

√ 60b(acω µε)3 µr = 2 3 3 2 3 3 πRs (2p a b + 2bc + p a c + ac ) εr

(5.5.2)

The permittivity and permeability of the dielectric filling are given by ε and µ, respectively, ω is angular frequency, and Rs is the surface resistivity of walls, which is related to the skin depth δs and the conductivity σ as follows. 1 = Rs = σδs

ωµ 2σ

(5.5.3)

On the other hand, if the cavity is filled with a dielectric with its loss tangent as tan δ while its walls are made of a perfect conductor, the quality factor Qd is given as 1 Qd = (5.5.4) tan δ

179

MICROWAVE RESONATORS

When there is power loss in both the cavity walls and the dielectric filling, the quality factor Q is found from Qc and Qd as follows: 1 1 1 + = Q Qc Qd

(5.5.5)

Example 5.7 An air-filled rectangular cavity is made from a piece of copper WR-90 waveguide. If it resonates at 9.379 GHz in TE101 mode, find the required length c and the Q value of this resonator. SOLUTION Specifications of WR-90 are given in Table A3.3 as follows: a = 0.9 in. = 2.286 cm and b = 0.4 in. = 1.016 cm From (5.5.1), 1 = 2c

9379 300

2 −

1 2a

2 = 22.3383 ⇒ c = 2.238 cm

Next, the surface resistance, Rs , is determined from (5.5.3) as 0.0253 , and the Qc is found from (5.5.2) to be about 7858. Example 5.8 A rectangular cavity made of copper has inner dimensions a = 1.6 cm, b = 0.71 cm, and c = 1.56 cm. It is filled with Teflon (εr = 2.05 and tan δ = 2.9268 × 10−4 ). Find the TE101 mode resonant frequency and Q value of this cavity. SOLUTION From (5.5.1), 300 fr = √ 2.05

1 2 × 0.016

2 +

1 2 × 0.0156

2 ≈ 9379 MHz = 9.379 GHz

Since power dissipates both in the dielectric filling as well as in the sidewalls, overall Q will be determined from (5.5.5). From (5.5.2) and (5.5.4), Qc and Qd are found to be 5489 and 3417, respectively. Substituting these into (5.5.5), the Q value of this cavity is found to be 2106. Circular Cylindrical Cavities Figure 5.22 shows the geometry of a circular cylindrical cavity of radius r and height h. It is filled with a dielectric material of relative permeability µr and

180

RESONANT CIRCUITS z r

h y

x

Figure 5.22 Geometry of the circular cylindrical cavity resonator.

dielectric constant εr . Its resonant frequency fr in megahertz is given as χnm 2 pπ 2 300 fr (MHz) = + MHz (5.5.6) √ 2π µr εr r h

where χnm =

pnm p nm

for TM modes for TE modes

(5.5.7)

As discussed in Section 4.9, pnm represents the mth zero of the Bessel function of the first kind and order n and p nm represents the mth zero of the derivative of the Bessel function. In other words, these represent roots of the following equations, respectively (see Table 5.3): Jn (x) = 0

(5.5.8)

dJn (x) =0 dx

(5.5.9)

and

The Qc of a circular cylindrical cavity operating in TEnmp mode and filled with a lossless dielectric can be found from the following formula: 2 2 1.5 1 − n/pnm (pnm ) + (pπr/ h)2 47.7465 Qc = δs fr (MHz) p 2 + (2r/ h) (pπr/ h)2 + 1 − (2r/ h) npπr/p h2 nm

nm

(5.5.10)

TABLE 5.3 Zeros of Jn (x ) and Jn (x )

n

pn1

pn2

pn3

pn1

pn2

pn3

0 1 2

2.405 3.832 5.136

5.520 7.016 8.417

8.654 10.173 11.620

3.832 1.841 3.054

7.016 5.331 6.706

10.173 8.536 9.969

181

MICROWAVE RESONATORS

In the case of TMnmp mode, Qc is given by 47.7465 Qc = δs fr (MHz) For p = 0, Qc ≈

2 + (pπr/ h)2 pnm 1 + (2r/ h)

47.7465 pnm δs fr (MHz) 1 + (r/ h)

p>0

p=0

(5.5.11)

(5.5.12)

The quality factor Qd due to dielectric loss and the overall Q are determined as before from (5.5.4) and (5.5.5) with appropriate Qc . Example 5.9 Determine the dimensions of an air-filled circular cylindrical cavity that resonates at 5 GHz in TE011 mode. It should be made of copper, and its height should be equal to its diameter. Find the Q of this cavity given that n = 0, m = p = 1, and h = 2r. SOLUTION From (5.5.6), with χ01 as 3.832 from Table 5.3, we get 3.8322 + (π/2)2 r= = 0.03955 m (5000 × 2π/300)2 and h = 2r = 0.0791 m Since δs =

2 = ωµσ

2π × 5 ×

109

2 = 9.3459 × 10−7 m × 4π × 10−7 × 5.8 × 107

The Q value of the cavity can be found from (5.5.10) as Qc =

47.7465 × 106 [3.8322 + (π/2)2 ]1.5 × = 39, 984.6 9.3459 × 10−7 × 5 × 109 [3.8322 + (π/2)2 ] + 1

Further, there is no loss of power in air (tan δ ≈ 0), Qd is infinite, and therefore Q is the same as Qc . Dielectric Resonators Dielectric resonators (DRs), made of high-permittivity ceramics, provide high Q values with smaller size. These resonators are generally a solid sphere or cylinder of circular or rectangular cross section. High-purity TiO2 (εr ≈ 100, tan δ ≈ 0.0001) was used in early dielectric resonators. However, it was found

182

RESONANT CIRCUITS

h

2r

Figure 5.23 Geometry of an isolated circular cylindrical dielectric resonator.

to be intolerably dependent on the temperature. With the development of new ceramics, temperature dependence of DR can be reduced significantly. Figure 5.23 shows an isolated circular cylindrical DR. Its analysis is beyond the scope of this book. Its TE01δ -mode resonant frequency, fr in GHz, can be found from the following formula: fr (GHz) =

r 34 3.45 + rεr h

GHz

(5.5.13)

where r and h are in millimeters. This relation is found to be accurate within ±2% if r > 0.5 h

(5.5.14)

50 > εr > 30

(5.5.15)

2> and

An isolated DR is of almost no use in practice. A more practical situation is illustrated in Figure 5.24, where it is coupled with a microstrip line in a conducting enclosure. Assume that the dielectric constants of the DR and the substrate are εr and εs , respectively. With various dimensions as shown (all in meters), the following formulas can be used to determine its radius r and the height h. If r is selected between the following bounds, the formulas presented in this section are found to have a tolerance of no more than 2% (Kajfez and Guillon, 1986): 1.2892 × 108 1.2892 × 108 >r> (5.5.16) √ √ fr εs fr εr where fr is resonant frequency in hertz. The height of the DR is then found as α1 α2 1 −1 −1 tan + tan h= β β tanh α1 t β tanh α2 d

(5.5.17)

183

MICROWAVE RESONATORS

where k 2 − k02 εs α2 = k 2 − k02 β = k02 εr − k 2 α1 =

(5.5.18) (5.5.19) (5.5.20)

y 2.405 + r 2.405r[1 + (2.43/y) + 0.291y] y = (k0 r)2 (εr − 1) − 5.784

k =

and

√ k0 = ω µ0 ε0

(5.5.21) (5.5.22)

(5.5.23)

Example 5.10 Design a TE01δ -mode cylindrical dielectric resonator for use at 35 GHz in the microstrip circuit shown in Figure 5.24. The substrate is 0.25 mm thick and its dielectric constant is 9.9. The dielectric constant of the material available for DR is 36. Further, the top of the DR should have a clearance of 1 mm from the conducting enclosure. SOLUTION Since 1.2892 × 108 = 1.171 × 10−3 m √ fr εs and 1.2892 × 108 = 6.139 × 10−4 m √ fr εr the radius r of the DR may be selected as 0.835 mm. Conducting enclosure d Microstrip

DR

h Substrate

2r

t

Figure 5.24 Geometry of a circular cylindrical dielectric resonator in MIC configuration.

184

RESONANT CIRCUITS

The height h of the DR is then determined from (5.5.17) to (5.5.22), as follows: y = (k0 r)2 (εr − 1) − 5.784 = 2.707 2.405 y + = 3.382 × 103 r 2.405r[1 + (2.43/y) + 0.291y] α1 = k 2 − k02 εs = 2.474 × 103 α2 = k 2 − k02 = 3.302 × 103 β = k02 εr − k 2 = 2.812 × 103

k =

and

After substituting these numbers into (4.5.17), h is found to be α1 α2 1 h= tan−1 + tan−1 = 6.683 × 10−4 m = 0.668 mm β β tanh α1 t β tanh α2 d

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Kajfez, D., and P. Guillon, Dielectric Resonators. Dedham, MA: Artech House, 1986. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P.A., Microwave Engineering. Englewood Cliffs, NJ: Prentice Hall, 1988. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1998.

PROBLEMS 5.1. Determine the resonant frequency, bandwidth, and Q of the circuit shown in Figure P5.1. 2 nH

0.015 µF

vin(t)

50 Ω

Figure P5.1

vo(t)

185

PROBLEMS

5.2. The quality factor of a 100-nH inductor is 150 at 100 MHz. It is used in a series resonant circuit with a 50- load resistor. Find the capacitor required to resonate this circuit at 100 MHz. Find the loaded Q value of the circuit. 5.3. Find the capacitance C in the series R –L–C circuit shown in Figure P5.3 if it is resonating at 1200 rad/s. Compute the output voltage vo (t) when vin (t) = 4 cos(ωt + 0.2π) V for ω as 1200, 300, and 4800 rad/s. 10 mH

C

vin(t)

2Ω

vo(t)

Figure P5.3

5.4. The magnitude of input impedance in the series R –L–C circuit shown in Figure P5.4 is found to be 3 at a resonant frequency of 1600 rad/s. Find the inductance L, Q, and the bandwidth.

L

20 µF

vin(t)

vo(t) R

Figure P5.4

5.5. Determine the element values of a resonant circuit to filter all the sinusoidal signals from 100 to 130 MHz. This filter is to be connected between a voltage source with negligible internal impedance and a communication system that has an input impedance of 50 . Plot its response over the frequency range 50 to 200 MHz. 5.6. A series resonant circuit is resonating at 105 rad/s and has a bandwidth of 100 rad/s. If the resistance in the circuit is 50 , find its inductance and capacitance. 5.7. For the circuit given in Figure P5.7, the poles are located as illustrated in Figure P7.2. Find (a) the Q, (b) ω0 , (c) ω, (d) ω1 , and (e) ω2 .

186

RESONANT CIRCUITS

jω

L

C 98,000 −σ

vin(t)

50 Ω

−98,000

vo(t)

−2 × 104 (a)

(b)

Figure P5.7 (a) Series R –L–C circuit with input–output terminals; (b) poles of the transfer function.

5.8. Calculate the resonant frequency, Q, and bandwidth of the parallel resonant circuit shown in Figure P5.8.

2 nH

0.015 µF

100 kΩ

Figure P5.8

5.9. Find the resonant frequency, unloaded Q, and loaded Q of the parallel resonant circuit shown in Figure P5.9.

500 Ω 50 pF

20 nH Resonator circuit

1 kΩ

Load

Figure P5.9

5.10. The parallel R –L–C circuit shown in Figure P5.10 is resonant at 20,000 rad/s. If its admittance has a magnitude of 1 mS at the resonance, find the values of R and L.

187

PROBLEMS

L

0.05 µF

R

Figure P5.10

5.11. Consider the loaded parallel resonant circuit shown in Figure P5.11. Compute the resonant frequency in radians per second, unloaded Q, and loaded Q of this circuit.

50 µF

Load 20 mH

2 kΩ

10 kΩ

Figure P5.11

5.12. Resistance R in a parallel R –L–C circuit is 200 . The circuit has a bandwidth of 80 rad/s with a lower half-power frequency at 800 rad/s. Find the inductance and capacitance of the circuit. 5.13. A parallel R –L–C circuit is resonant at 2 × 106 rad/s and has a bandwidth of 20,000 rad/s. If its impedance at resonance is 2000 , find the circuit parameters. 5.14. A resonator is fabricated from a 2λ-long transmission line that is shortcircuited at its ends. Find its Q. 5.15. A 3-cm-long 100- air-filled coaxial line is used to fabricate a resonator. It is short-circuited at one end while a capacitor is connected at the other end to obtain the resonance at 6 GHz. Find the value of this capacitor. 5.16. A 100- air-filled coaxial line of length l is used to fabricate a resonator. It is terminated by 10 −j 5000 at its ends. If the signal wavelength is 1 m, find the required length for the first resonance and the Q of this resonator. 5.17. Design a half-wavelength-long coaxial line resonator that is short-circuited at its ends. Its inner conductor radius is 0.455 mm and the inner radius of the outer conductor is 1.499 mm. The conductors are made of copper. Compare the Q value of an air-filled resonator to that of a Teflon-filled resonator operating at 800 MHz. The dielectric constant of Teflon is 2.08 and its loss tangent is 0.0004. 5.18. Design a half-wavelength-long microstrip resonator using a 50- line that is short-circuited at its ends. The substrate thickness is 0.12 cm with a

188

RESONANT CIRCUITS

dielectric constant of 2.08 and a loss tangent of 0.0004. The conductors are of copper. Compute the length of the line for resonance at 900 MHz and the Q value of the resonator. 5.19. The secondary side of a tightly coupled transformer is terminated by a 2-k load in parallel with capacitor C2 . A current source I = 10 cos(4 × 106 ) mA is connected on its primary side. The inductance L1 of its primary is 0.30 µH, the mutual inductance M = 3 µH, and the unloaded Q of the secondary coil is 70. Determine the value of C2 such that this circuit resonates at 4 × 106 rad/s. Find its input impedance and the output voltage at the resonant frequency. What is the circuit bandwidth? 5.20. A 16- resistor terminates the secondary side of a tightly coupled transformer. A 40-µF capacitor is connected in series with its primary coil, which has an inductance L1 at 100 mH. If the secondary coil has an inductance L2 at 400 mH, find the resonant frequency and the Q value of the circuit. 5.21. A 16- resistor terminates the secondary side of a tightly coupled transformer. A 30-µF capacitor is connected in series with its primary coil, which has an inductance of 25 mH. If the circuit Q value is 50, find the inductance L2 of the secondary coil. 5.22. An air-filled rectangular cavity is made from a piece of copper WR-430 waveguide. If it resonates at 2 GHz in TE101 mode, find the required length c and the Q of this resonator. 5.23. Design an air-filled circular cylindrical cavity that resonates at 9 GHz in TE011 mode. It should be made of copper, and its height should be equal to its diameter. Find the Q value of this cavity. 5.24. Design a TE01δ mode cylindrical dielectric resonator for use at 4.267 GHz in the microstrip circuit shown in Figure 5.24. The substrate is 0.7 mm thick and its dielectric constant is 9.6. The dielectric constant of the material available for the DR is 34.19. Further, the top of the DR should have a clearance of 0.72 mm from the conducting enclosure.

6 IMPEDANCE-MATCHING NETWORKS

One of the most critical requirements in the design of high-frequency electronic circuits is that the maximum possible signal energy is transferred at each point. In other words, the signal should propagate in a forward direction with a negligible echo (ideally, zero). Echo signal not only reduces the power available but also deteriorates the signal quality due to the presence of multiple reflections. As noted in Chapter 5, impedance can be transformed to a new value by adjusting the turns ratio of a transformer that couples it with the circuit. However, it has several limitations. In this chapter we present a few techniques to design other impedance-transforming networks. These circuits include transmission line stubs and resistive and reactive networks. Further, the techniques introduced are needed in active circuit design at RF and microwave frequencies. As shown in Figure 6.1, impedance-matching networks are employed at the input and output of an amplifier circuit. These networks may also be needed to perform other tasks, such as filtering the signal and blocking or passing the dc bias voltages. This chapter begins with a section on impedance-matching techniques that use a single reactive element or stub connected in series or in shunt. Theoretical principles behind the technique are explained, and the graphical procedure to design these circuits using a Smith chart is presented. Principles and procedures of double-stub matching are discussed in the following section. The chapter ends with sections on resistive and reactive L-section matching networks. Both analytical and graphical procedures to design these networks using ZY charts are included. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

189

190

Zs

Vs

IMPEDANCE-MATCHING NETWORKS

Impedancetransforming circuit

Figure 6.1

Transistor with dc bias circuit

Impedancetransforming circuit

Load ZL

Block diagram of an amplifier circuit.

6.1 SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS When a lossless transmission line is terminated by an impedance ZL , the magnitude of the reflection coefficient (and hence, the VSWR) on it remains constant, but its phase angle can be anywhere between +180◦ and −180◦ . As we have seen in Chapter 3, it represents a circle on a Smith chart, and a point on this circle represents the normalized load. As one moves away from the load, the impedance (or the admittance) value changes. This movement is clockwise on the VSWR circle. The real part of the normalized impedance (or normalized admittance) becomes unity at certain points on the line. Addition of a single reactive element or a transmission line stub at this point can eliminate the echo signal and reduce the VSWR to unity beyond this point. A finite-length transmission line with its other end an open or short circuit is called a stub and behaves like a reactive element, as explained in Chapter 3. In this section we discuss the procedure for determining the location on a lossless feeding line where a stub or a reactive element can be connected to eliminate the echo signal. Two different possibilities, a series or a shunt element, are considered. Mathematical equations as well as graphical methods are presented to design the circuits.

Shunt Stub or Reactive Element Consider a lossless transmission line of characteristic impedance Z0 that is terminated by a load admittance YL , as shown in Figure 6.2. Corresponding normalized input admittance at a point ds away from the load can be found from (3.2.6) as Y L + j tan βds Y in = (6.1.1) 1 + j Y L tan βds To obtain a matched condition at ds , the real part of the input admittance must be equal to the characteristic admittance of the line [i.e., the real part of (6.1.1) must be unity]. This requirement is used to determine ds . The parallel susceptance Bs is then connected at ds to cancel out the imaginary part of Yin .

191

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

ds

YL = GL + jBL

jBs

Figure 6.2

Transmission line with a shunt matching element.

Hence, ds =

1 tan−1 β

2 B L ± B L − A(1 − GL ) A

(6.1.2)

2

where A = GL (GL − 1) + B L . The imaginary part of the normalized input admittance at ds is found as follows: 2

B in =

(B L + tan βds )(1 − B L tan βds ) − GL tan βds (GL tan βds )2 + (1 − B L tan βds )2

(6.1.3)

The other requirement to obtain a matched condition is B s = −B in

(6.1.4)

Hence, a shunt inductor is needed at ds if the input admittance is found capacitive (i.e., Bin is positive). On the other hand, it will require a capacitor if Yin is inductive at ds . As mentioned earlier, a lossless transmission line section can be used in place of this inductor or capacitor. The length of this transmission line section is determined according to the susceptance needed by (6.1.4) and the termination (i.e., an open circuit or a short circuit) at its other end. This transmission line section is called a stub. If ls is the stub length that has a short circuit at its other end, ls =

1 1 cot−1 (−B s ) = cot−1 B in β β

(6.1.5)

On the other hand, if there is an open circuit at the other end of the stub, ls =

1 1 tan−1 B s = tan−1 (−B in ) β β

(6.1.6)

Series Stub or Reactive Element If a reactive element (or stub) needs to be connected in series as shown in Figure 6.3, the design procedure can be developed as follows. The normalized

192

IMPEDANCE-MATCHING NETWORKS

ds

jXs ZL = RL + jXL

Figure 6.3 Transmission line with a matching element connected in series.

input impedance at ds is Z in =

Z L + j tan βds

(6.1.7)

1 + j Z L tan βds

To obtain a matched condition at ds , the real part of the input impedance must be equal to the characteristic impedance of the line [i.e., the real part of (6.1.7) must be unity]. This condition is used to determine ds . A reactance Xs is then connected in series at ds to cancel out the imaginary part of Zin . Hence,

ds =

1 tan−1 β

XL ±

2 X L − Az (1 − R L ) Az

(6.1.8)

2

where Az = R L (R L − 1) + XL . The imaginary part of the normalized input impedance at ds is found as follows: 2

X in =

(X L + tan βds )(1 − X L tan βds ) − R L tan βds (R L tan βds )2 + (1 − X L tan βds )2

(6.1.9)

To obtain a matched condition at ds , the reactive part Xin must be eliminated by adding an element of opposite nature. Hence, Xs = −Xin

(6.1.10)

Therefore, a capacitor will be needed in series if the input impedance is inductive. It will require an inductor if the input reactance is capacitive. As before, a transmission line stub can be used instead of an inductor or a capacitor. The length of this stub with an open circuit at its other end can be determined as follows: ls =

1 1 cot(−X s ) = cot Xin β β

(6.1.11)

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

193

However, if the stub has a short circuit at its other end, its length will be a quarter-wavelength shorter (or longer, if the resulting number becomes negative) than this value. It can be found as ls =

1 1 tan Xs = tan(−X in ) β β

(6.1.12)

Note that the location ds and the stub length ls are periodic in nature in both cases. It means that the matching conditions will also be satisfied at points one-half wavelength apart. However, the shortest possible values of ds and ls are preferred because those provide the matched condition over a broader frequency band. Graphical Method These matching networks can also be designed graphically using a Smith chart. The procedure is similar for both series- and shunt-connected elements, except that the former is based on the normalized impedance, whereas the latter works with normalized admittance. It can be summarized in the following steps: 1. Determine the normalized impedance of the load and locate that point on a Smith chart. 2. Draw the constant-VSWR circle. If the stub needs to be connected in parallel, move a quarter-wavelength away from the load impedance point. This point is located at the other end of the diameter that connects the load point with the center of the circle. For a series stub, stay at the normalized impedance point. 3. From the point found in step 2, move toward the generator (i.e., clockwise) on the VSWR circle until it intersects the unity resistance (or conductance) circle. The distance traveled to this intersection point from the load is equal to ds . There will be at least two such points within one-half wavelength from the load. A matching element can be placed at either one of these points. 4. If the admittance in step 3 is 1 + j B, a susceptance of −j B in shunt is needed for matching. This can be a discrete reactive element (inductor or capacitor, depending on a negative or positive susceptance value) or a transmission line stub. 5. In the case of a stub, the length required is determined as follows. Since its other end will have an open or a short, the VSWR on it will be infinite. It is represented by the outermost circle of the Smith chart. Locate the desired susceptance point (i.e., 0 − j B) on this circle and then move toward load (counterclockwise) until an open circuit (i.e., a zero susceptance) or a short circuit (an infinite susceptance) is found. This distance is equal to the stub length ls . For a series reactive element or stub, steps 4 and 5 will be the same except that the normalized reactance replaces the normalized susceptance.

194

IMPEDANCE-MATCHING NETWORKS

Example 6.1 A uniform, lossless 100- line is connected to a load of 50 − j 75 , as illustrated in Figure 6.4. A single stub of 100- characteristic impedance is connected in parallel at a distance ds from the load. Find the shortest values of ds and stub length ls for a match. SOLUTION As mentioned in the preceding analysis, design equations (6.1.2), (6.1.3), (6.1.5), and (6.1.6) for a shunt stub use admittance parameters. On the other hand, the series-connected stub design uses impedance parameters in (6.1.8), (6.1.9), (6.1.11), and (6.1.12). Therefore, ds and ls can be determined theoretically, as follows: YL =

YL Z0 100 = = = 0.6154 + j 0.9231 Y0 ZL 50 − j 75 2

A = GL (GL − 1) + B L = 0.6154(0.6154 − 1) + 0.92312 = 0.6154 From (6.1.2), the two possible values of ds are λ (−0.75)2 − 0.6154(1 − 0.5) −1 −0.75 + ds = tan = 0.1949λ 2π 0.6154 and λ −0.75 − tan−1 ds = 2π

(−0.75)2 − 0.6154(1 − 0.5) = 0.0353λ 0.6154

At 0.1949λ from the load, the real part of the normalized admittance is unity and its imaginary part is −1.2748. Hence, the stub should provide j 1.2748 to cancel it out. Length of a short-circuited stub, ls , is calculated from (6.1.5) as ls =

1 cot−1 (−1.2748) = 0.3941λ β

ds

ZL = 50 − j75 Ω

ls

Figure 6.4 Shunt stub matching network.

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

195

On the other hand, normalized admittance is 1 + j 1.2748 at 0.0353λ from the load. To obtain a matched condition, the stub at this point must provide a normalized susceptance of −j 1.2748. Hence, ls =

1 cot−1 (1.2748) = 0.1059λ β

Thus, there are two possible solutions to this problem. In one case, a shortcircuited 0.3941λ-long stub is needed at 0.1949λ from the load. The other design requires a 0.1059λ-long short-circuited stub at 0.0353λ from the load. It is preferred over the former design because of its shorter lengths. The following steps are needed to solve this example graphically with a Smith chart: 1. Determine the normalized load admittance. ZL =

50 − j 75 = 0.5 − j 0.75 100

2. Locate the normalized load impedance point on the Smith chart. Draw the VSWR circle as shown in Figure 6.5. 3. From the load impedance point, move to the diametrically opposite point and locate the corresponding normalized load admittance. It is the point 0.62 + j 0.91 on the chart. 4. Locate the point on the VSWR circle where the real part of the admittance is unity. There are two such points with normalized admittance values 1 + j 1.3 (say, point A) and 1 − j 1.3 (say, point B), respectively. 5. The distance ds of 1 + j 1.3 (point A) from the load admittance can be determined as 0.036λ (i.e., 0.170λ − 0.134λ) and for point B (1 − j 1.3) as 0.195λ (i.e., 0.329λ − 0.134λ). 6. If a susceptance of −j 1.3 is added at point A or j 1.3 at point B, the load will be matched. 7. Locate the point −j 1.3 along the lower circumference of the chart and from there move toward the load (counterclockwise) until the short circuit (infinity on the chart) is reached. Separation between these two points is 0.25λ − 0.146λ = 0.104λ. Hence a 0.104λ-long transmission line with a short circuit at its rear end will have the desired susceptance for point A. 8. For a matching stub at point B, locate the point j 1.3 on the upper circumference of the chart and then move toward the load up to the short circuit (i.e., the right-hand end of the chart). Hence, the stub length ls for this case is determined as 0.25λ + 0.146λ = 0.396λ. Therefore, a 0.104λ-long stub at 0.036λ from the load (point A) or a 0.396λlong stub at 0.195λ (point B) from the load will match the load. These values are comparable to those obtained earlier.

196

1

0.8

0.15 0.35

2

0.4

0 0.2 0 0.3

S TOWARD GE N ENGTH VEL THS TOWARD LERATO -WA - AVELENG 0.0 OAD R -W - -> LEN 0.0 E V A W

-WA

4

0.

0 0.2 0 0.3

0.0 0.4 5 5

0.5

0.6

0.7

0.10 0.40

0.9

SINGLE REACTIVE ELEMENT OR STUB MATCHING NETWORKS

0.2

5

10 10

5

2

0.5

0.4

0.3

0.2

0.1

0.1

0.25 0.25

10

0.1

0.2

5 0.3 0.2 0 0

4

0.

1

0.9

0.8

0.7

0.40 0.10

Figure 6.10

0.35 0.15

0.6

2

0.5

5 0.4 5 0.0

0.3

Graphical solution to Example 6.3.

the corresponding input impedance point is identified as 1.27 + j 0.37. For the circuit given in Figure 6.9(a), the admittance (normalized) points are found as 0.45 + j 0.22 and 0.73 − j 0.21, respectively. Next, move from the normalized load admittance point toward the generator by a distance of 0.083583λ (i.e., 0.042λ + 0.084λ = 0.126λ of the scale “wavelengths toward generator”). The normalized admittance value of this point is found to be 0.73 + j 0.69. Hence, its real part is the same as that of the desired input admittance. However, its susceptance is j 0.69, whereas the desired value is −j 0.2. Hence, an inductor will be needed in parallel at this point. Since the given circuit has a capacitor, this design is not possible. For the circuit shown in Figure 6.9(b), elements are connected in series. Therefore, normalized impedance points need to be used in this case. Move from the load impedance (1.8 − j 0.85) point toward the generator by a distance of 0.029928λ (i.e., 0.292λ + 0.03λ = 0.322λ on the “wavelengths toward generator” scale). Normalized impedance value at this point is 1.27 − j 0.95. Thus,

202

IMPEDANCE-MATCHING NETWORKS

the resistance at this point is found to be equal to the value desired. However, its reactance is −j 0.95, whereas the required value is j 0.37. Therefore, a series reactance of j 1.32 is needed at this point. The given circuit has an inductor that provides a positive reactance. Hence, this circuit will work. The required inductance L is found as L=

1.32 × 50 H = 2.626 nH 2π × 4 × 109

6.2 DOUBLE-STUB MATCHING NETWORKS The matching technique presented in Section 5.1 requires that a reactive element or stub be placed at a precise distance from the load. This point will shift with load impedance. Sometimes it may not be feasible to match the load using a single reactive element. Another possible technique to match the circuit employs two stubs with a fixed separation between them. This device can be inserted at a convenient point before the load. The impedance is matched by adjusting the lengths of the two stubs. Of course, it does not provide a universal solution. As will be seen later in this section, separation between the two stubs limits the range of load impedance that can be matched with a given double-stub tuner. Let l1 and l2 be the lengths of two stubs, as shown in Figure 6.11. The first stub is located at a distance l from the load, ZL = RL + j XL ohms. Separation between the two stubs is d, and the characteristic impedance of every transmission line is Z0 . In double-stub matching, the load impedance ZL is transformed to the normalized admittance at the location of the first stub. Since the stub is connected in parallel, its normalized susceptance is added to that and the resulting normalized admittance is transferred to the location of the second stub. Matching conditions at this point require that the real part of this normalized admittance

l

d

Z0

Z0

l2

Figure 6.11

ZL = RL + jXL

l1

Double-stub matching network.

DOUBLE-STUB MATCHING NETWORKS

203

be equal to unity while its imaginary part is canceled by a conjugate susceptance of the second stub. Mathematically, Y + j (B 1 + tan βd) 1 + j (Y + j B 1 ) tan βd where Y =

1 + j Z L tan βl Z L + j tan βl

=

+ j B2 = 1

Y L + j tan βl 1 + j Y L tan βl

= G + jB

(6.2.1)

(6.2.2)

j B1 and j B2 are the susceptance of the first and second stubs, respectively, and β is the propagation constant over the line. For Re

Y + j (B 1 + tan βd) 1 + j (Y + j B 1 ) tan βd

=1

2

G tan2 βd − G(1 + tan2 βd) + [1 − (B + B 1 ) tan βd]2 = 0

(6.2.3)

Since conductance of the passive network must be a positive quantity, (6.2.3) requires that a given double stub can be used for matching only if the following condition is satisfied: 0 ≤ G ≤ csc2 βd (6.2.4) Two possible susceptances of the first stub that can match the load are determined by solving (6.2.3):

2 2 B 1 = cot βd 1 − B tan βd ± G sec βd − (G tan βd)

(6.2.5)

Normalized susceptance of the second stub is determined from (6.2.1) as 2

B2 =

G tan βd − (B + B 1 + tan βd)[1 − (B + B 1 ) tan βd] (G tan βd)2 + [1 − (B + B 1 ) tan βd]2

(6.2.6)

Once the susceptance of a stub is known, its short-circuit length can be determined easily as follows: 1 l1 = cot−1 (−B 1 ) (6.2.7) β and l2 =

1 cot−1 (−B 2 ) β

(6.2.8)

204

IMPEDANCE-MATCHING NETWORKS

Graphical Method A two-stub matching network can also be designed graphically with the help of a Smith chart. This procedure follows the analytical concepts discussed above and can be summarized as follows: 1. Locate the normalized load-impedance point on a Smith chart and draw the VSWR circle. Move to the corresponding normalized admittance point. If the load is connected right at the first stub, go to the next step; otherwise, move toward the generator (clockwise) by 2βl on the VSWR circle. Assume that the normalized admittance of this point is g + j b. 2. Rotate the unity-conductance circle counterclockwise by 2βd. The conductance circle that touches this circle encloses the “forbidden region.” In other words, this tuner can match only those Y that lie outside this circle. It is a graphical representation of the condition expressed by (6.2.4). 3. From g + j b, move on the constant-conductance circle until it intersects the rotated unity-conductance circle. There are at least two such points, providing two design possibilities. Let the normalized admittance of one of these points be g + j b1 . 4. The normalized susceptance required of the first stub is j (b1 − b). 5. Draw a VSWR circle through point g + j b1 and move toward the generator (clockwise) by 2βd on it. This point will fall on the unity-conductance circle of the Smith chart. Assume that this point is 1 + j b2 . 6. The susceptance required from the second stub is −j b2 . 7. Once the stub susceptances are known, their lengths can be determined following the procedure used in the technique described earlier. Example 6.4 For the double-stub tuner shown in Figure 6.12, find the shortest values of l1 and l2 to match the load.

λ/8

λ/8 Z0 = 50 Ω

Z0 = 50 Ω

Z0 = 50 Ω

ZL = 100 + j50 Ω Z0 = 50 Ω

l2

l1

Figure 6.12 Two-stub matching network for Example 6.4.

205

DOUBLE-STUB MATCHING NETWORKS

SOLUTION Since the two stubs are separated by λ/8, βd is equal to π/4 and the condition (6.2.4) gives 0≤G≤2 This means that the real part of the normalized admittance at the first stub (loadside stub) must be less than 2; otherwise, it cannot be matched. The graphical procedure requires the following steps to find stub settings: 100 + j 50 = 2 + j1 50 2. Locate the normalized load impedance on a Smith chart (point A in Figure 6.13) and draw the VSWR circle. Move to the diametrically opposite side and locate the corresponding normalized admittance point B at 0.4 − j 0.2.

0.8

1

0.15 0.35

2

F 4

0.

0 0.2 0 0.3

0.0 0.4 5 5

0.5

0.6

0.7

0.10 0.40

0.9

1. Z L =

0.2

5

E A

5

2

10

0.5

0.4

0.3

0.2

0.1

0.1

D

10 0.25 0.25

OWARD GENERA GTHS T TO LEN E S TOWARD LOAD R ---> V H T -WA VELENG A 0.0 -W 10 ), Rp is 50 while Rs is 10 . Therefore, Q2 =

50 −1=4⇒Q=2 10

214

IMPEDANCE-MATCHING NETWORKS

Xs and Xp can now be determined from (6.3.8) and (6.3.9): Xs = QRs = 2 × 10 = 20 and Xp =

50 Rp = = 25 Q 2

Reactance Xp is connected in parallel with the transmission line and Xs is in series with the load. These two matching circuits are shown in Figure 6.20. If Xp is a capacitive reactance of 25 , Xs must be inductive 20 . The reactive part that has not been taken into account so far needs to be included at this point in Xs . Since this reactive part of the load is inductive 10 , another series inductor for the remaining 10 is needed, as shown in circuit (1) of Figure 6.20. Hence, ωLs = 20 − 10 Therefore, Ls =

10 H = 0.3183 × 10−8 H = 3.183 nH 2π × 500 × 106

and

1 = 25 ωCp

Therefore, Cp =

1 = 0.0127324 × 10−9 F = 12.7324 pF 2π × 500 × 106 × 25

In the second case, Xp is assumed inductive and therefore Xs needs to be capacitive. It is circuit (2) in Figure 6.20. Since the load has a 10- inductive reactance,

Cs

Ls

10 Ω Cp 50 Ω

Load

10 Ω 50 Ω

Lp

j10 Ω

j10 Ω (1)

Load

(2)

Figure 6.20 Reactive matching circuits for Example 6.8.

MATCHING NETWORKS USING LUMPED ELEMENTS

215

it must be taken into account in determining the required capacitance. Therefore, ωCs =

1 1 ⇒ Cs = F = 10.61 pF 20 + 10 2π × 500 × 106 × 30

and ωLp = 25 ⇒ Lp =

25 H = 7.9577 nH 2π × 500 × 106

The matching procedure used so far is based on the transformation of series impedance to a parallel circuit of the same quality factor. Similarly, one can use an admittance transformation, as shown in Figure 6.21. These design equations can be obtained easily following the procedure used earlier: Gs = 1 + Q2 Gp

(6.3.11)

Q=

Gs Bs

(6.3.12)

Q=

Bp Gp

(6.3.13)

and

Note from (6.3.11) that Gs must be greater than Gp for a real Q. Example 6.9 A load admittance, YL = 8 − j 12 mS, is to be matched with a 50- line. There are three different L-section matching networks given to you, as shown in Figure 6.22. Complete or verify each of these circuits. Find the element values at 1 GHz. SOLUTION Since the characteristic impedance of the line is 50 , the corresponding conductance will be 0.02 S. The real part of the load admittance is 0.008 S. Therefore, Gs must be 0.02 S in (6.3.11). However, the circuits in Figure 6.22 have a capacitor or an inductor connected in parallel with the 50- line. Obviously, these design equations cannot be used here. Equations (6.3.8)

Gs

Gp jBs

jBp

Figure 6.21 Series- and parallel-connected admittance circuits.

216

IMPEDANCE-MATCHING NETWORKS Cs

Ls

Load

Load Cp 50 Ω

Lp

50 Ω

YL

(a)

YL

(b) Cs

Load

Cp 50 Ω

YL

(c)

Figure 6.22 Three possible L-section matching circuits given in Example 6.9.

to (6.3.10) use the impedance values. Hence, we follow the procedure outlined below: YL =

1 103 ⇒ ZL = = 38.4615 + j 57.6923 ZL 8 − j 12

Therefore, Rp = 50 and Rs = 38.4615 ⇒ Q =

50 − 1 = 0.5477 38.4615

Now, from (6.3.8) and (6.3.9), Xs = QRs = 0.5477 × 38.4615 = 21.0654 and Xp =

50 Rp = = 91.2909 Q 0.5477

For circuit (a) in Figure 6.22, capacitor Cp can be selected, which provides 91.29 reactance. It means that the inductive reactance on its right-hand side must be 21.06 . However, an inductive reactance of 57.69 is already present there, due to load. Another series inductor will increase it further, whereas it needs to be reduced to 21.06 . Thus, we conclude that circuit (a) cannot be used.

217

MATCHING NETWORKS USING LUMPED ELEMENTS

In circuit (b) in Figure 6.22, there is an inductor in parallel with the transmission line. For a match, its reactance must be 91.29 , and overall reactance to the right must be capacitive 21.06 . Inductive reactance of 57.69 of the load needs to be neutralized as well. Hence, Xc = 57.6923 + 21.0654 = 78.7577 Therefore, C=

2π ×

109

1 F = 2.0208 pF × 78.7577

and Xp = 91.2909 ⇒ L =

91.2909 H = 14.5294 nH 2π × 109

Circuit (c) in Figure 6.22 has a capacitor across the transmission line terminals. Assuming that its reactance is 91.29 , reactance to the right must be inductive 21.06 . As noted before, the reactive part of the load is an inductive 57.69 . It can be reduced to the desired value by connecting a capacitor in series. Hence, the component values for circuit (c) are calculated as follows: Xs = 57.6923 − 21.0654 = 36.6269 Therefore, Cs = 4.3453 pF and Xp = 91.2909 ⇒ Cp = 1.7434 pF Example 6.10 Reconsider Example 6.9, where YL = 8 −j 12 mS is to be matched with a 50- line. Complete or verify the L-section matching circuits shown in Figure 6.23 at a signal frequency of 1 GHz.

Ls Cs

Cp 50 Ω

YL

(a)

Figure 6.23

Load

Load 50 Ω

Cp

YL

(b)

L-section matching networks given in Example 6.10.

218

IMPEDANCE-MATCHING NETWORKS

SOLUTION As noted in Example 6.9, Gs is 0.02 S and Gp is 8 mS. Looking over the given circuit configurations, it seems possible to complete or verify these circuits through (6.3.11) to (6.1.13). Hence, Gs 0.02 Q= − 1 = 1.2247 −1= Gp 0.008 Bs =

Gs 0.02 = = 0.0163 S Q 1.2247

and Bp = QGp = 1.2247 × 0.008 = 0.0097978 ≈ 0.01 S If Bs is selected as an inductive susceptance, Bp must be capacitive. Since load has an inductive susceptance of 12 mS, the capacitor must neutralize that, too. Hence, circuit (a) of Figure 6.23 will work provided that Bs = 0.0163 S ⇒ Ls = 9.746 nH and Bp = (0.01 + 0.012) S = 0.022 S ⇒ Cp = 3.501 pF In circuit (b) of Figure 6.23, a capacitor is connected in series with Gs . Therefore, Bp must be an inductive susceptance. Since it is 0.012 S whereas only 0.01 S is required for matching, a capacitor in parallel is needed. Hence, Bs = 0.0163 S ⇒ Cs = 2.599 pF and Bp = (0.012 − 0.01) = 0.002 S (capacitive) ⇒ Cp = 0.3183 pF Thus, both circuits can work provided that the component values are as found above. Graphical Method As described earlier, L-section reactive networks can be used for matching the impedance. One of these reactive elements appears in series with the load or the desired impedance, while the other one is connected in parallel. Thus, the resistive part stays constant when a reactance is connected in series with impedance. Similarly, a change in the shunt-connected susceptance does not affect the conductive part of the admittance. Consider the normalized impedance point X on the Smith chart shown in Figure 6.24. To distinguish it from others, it may be called the Z-Smith chart. If its resistive part needs to be kept constant at 0.5, one must stay on this constant-resistance circle. A clockwise movement from

219

MATCHING NETWORKS USING LUMPED ELEMENTS

0

jX

12

110

90

100

80

70 60 50

30

40

14 0

1

Inductive reactance

15 0

Capacitive reactance 170

10

0 10

5

2

1

0.5

0.2

−170

−16

−20

0

−10

Capacitive reactance

0

180

Inductive reactance

20

160

30

X

−

0 15

−3 0

−

0

0 14

−4 −5

0

30

−6

0

−70

20

−80

−1

0 − 1

−90

−100 −11

-jX

Figure 6.24 Impedance (Z-) Smith chart.

this point increases the positive reactance. It means that the inductance increases in series with the impedance of X. On the other hand, the positive reactance decreases with a counterclockwise movement. A reduction in positive reactance means a decrease in series inductance or an increase in series capacitance. Note that this movement also represents an increase in negative reactance. Now consider a Smith chart that is rotated by 180◦ from its usual position, as shown in Figure 6.25. It may be called a Y -Smith chart because it represents the admittance plots. In this case, addition (or subtraction) of a susceptance to admittance does not affect its real part. Hence, it represents a movement on the constant-conductance circle. Assume that a normalized admittance is located at point X. The conductive part of this admittance is 0.5. If a shunt inductance is added, it moves counterclockwise on this circle. On the other hand, a capacitive susceptance moves this point clockwise on the constant-conductance circle. A superposition of Z- and Y -Smith charts is shown in Figure 6.26. It is generally referred to as a ZY -Smith chart because it includes impedance as well as admittance plots at the same time. A short-circuit impedance is zero while the corresponding admittance goes to infinity. A single point on the ZY -Smith chart represents it. Similarly, it may be found that other impedance points of the Z chart coincide with the corresponding admittance points of the Y chart as well. Hence, impedance

220

IMPEDANCE-MATCHING NETWORKS

12 0

−100

−90

−80

−70

−6

0

−

−5

0

−1

4 −1

0

30

-110

−4 0

-jB

X

−10

0

0.2

0.5

1

2

5

0 10

Inductive susceptance

180

Capacitive susceptance

−170

−20

1 0 − −16

50

−3 0

Inductive susceptance

170

10

20

160

Capacitive susceptance

15

0

30 14

0

40 50

0

13

60

70

20

80

Figure 6.25

90

100

110

jB

1

Admittance (Y -) Smith chart..

can be transformed to corresponding admittance just by switching from the Z- to the Y -scales of a ZY -Smith chart. For example, a normalized impedance point of 0.9 − j 1 is located using the Z-chart scales as A in Figure 6.26. The corresponding admittance is read from a Y -chart as 0.5 + j 0.55. Reactive L-section matching circuits can be designed easily using a ZY -Smith chart. Load and desired impedance points are identified on this chart using Zscales. Y -scales may be used if either one is given as admittance. Note that the same characteristic impedance is used for normalizing these values. Now, move from the point to be transformed toward the desired point following a constantresistance or constant-conductance circle. Movement on the constant-conductance circle gives the required susceptance (i.e., a reactive element will be needed in shunt). It is determined by subtracting initial susceptance (the starting point) from the value reached. When moving on the conductance circle, use the Y -scale of the chart. Similarly, moving on a constant-resistance circle will mean that a reactance must be connected in series. It can be determined using the Z-scale of the ZY -Smith chart.

221

MATCHING NETWORKS USING LUMPED ELEMENTS

110

90

100

80

70 60

0 12

50

0

10

170

20

160

15

30

0

40

14 0

13

10

5

2

1

0.5

−20

0 −16

−10

−170

0.2

0

180

0

−

0

−4

0 14

0

−1

50

−3

A

−5

30

0

−6

0

20

−70

−80

−90

−100

0

−11

−1

−1

Figure 6.26 ZY -chart.

Note from the ZY -chart that if a normalized load value falls inside the unity resistance circle, it can be matched with the characteristic impedance Z0 only after shunting it with an inductor or capacitor (a series-connected inductor or capacitor at the load will not work). Similarly, a series inductor or capacitor is required at the load if the normalized value is inside the unity conductance circle. Outside these two unity circles, if the load point falls in the upper half, the first component required for matching would be a capacitor that can be series or parallel connected. On the other hand, an inductor will be needed at the load first if the load point is located in the lower half and outside the unity circles. Example 6.11 Use a ZY -Smith chart to design the matching circuits of Examples 6.9 and 6.10.

222

IMPEDANCE-MATCHING NETWORKS

SOLUTION In this case the given load admittance can be normalized by the characteristic admittance of the transmission line. Hence, YL =

YL = YL Z0 = 0.4 − j 0.6 Y0

This point is found on the ZY -Smith chart as A in Figure 6.27. The matching requires that this admittance must be transformed to 1. This point can be reached through a unity conductance or resistance circle. Hence, a matching circuit can be designed if somehow we can reach one of these circles through constantresistance and constant-conductance circles only. Since constant-conductance circles intersect constant- resistance circles, one needs to start on a constant-

110

90

100

80

70 60

0 12

50

0

14

40

0

13

0 15

30

A

20

160

B

10

170

D

10

5

2

1

0.5

O

−10

−170

0.2

0

180

0

−20

0 −16

E

−4 0

40 −1

−

0 15

−3 0

C

−5

30

0

−6

0

20

−70

−80

−90

−100

−11 0

−1

−1

Figure 6.27 Graphical solution to Example 6.11.

MATCHING NETWORKS USING LUMPED ELEMENTS

223

resistance circle to get on the unity conductance circle, or vice versa. There are two circles (i.e., 0.4 conductance and 0.78 resistance circles) passing through point A. Hence, it is possible to get on to the unity resistance circle via the 0.4 conductance circle. Alternatively, one can reach on the unity conductance circle through the 0.78 resistance circle. In either case, a circuit can be designed. From point A, one can move on the 0.4 conductance circle (or 0.78 resistance circle) so that the real part of the admittance (or impedance) remains constant. If we start counterclockwise from A on the conductance circle, we end up at infinite susceptance without intersecting the unity resistance circle. Obviously, this does not yield a matching circuit. Similarly, a clockwise movement on the 0.78 resistance circle from A cannot produce a matching network. Hence, an inductor in parallel or in series with the given load cannot be used for the design of a matching circuit. This proves that the circuit shown in Figure 6.22(a) cannot be designed. There are four possible circuits. In one case, move from point A to B on the conductance circle (a shunt capacitor) and then from B to O on the unity resistance circle (a series capacitor). The second possibility is via A to C (a shunt capacitor) and then from C to O (a series inductor). A third circuit can be obtained following the path from A to D (a series capacitor) and then from D to O (a shunt capacitor). The last circuit can be designed by following the path from A to E (a series capacitor) and then from E to O (a shunt inductor). A–E to E–O and A–D to D–O correspond to the circuits shown in Figure 6.22(b) and (c), respectively. Similarly, A–C to C–O and A–B to B–O correspond to circuits shown in Figure 6.23(a) and (b), respectively. Component values for each of these circuits are determined following the corresponding susceptance or reactance scales. For example, we move along a resistance circle from A to D. That means that there will be a change in reactance. The required element value is determined after subtracting the reactance at A from the reactance at D. This difference is j 0.42 − j 1.15 = −j 0.73. Negative reactance means that it is a series capacitor. Note that −j 0.73 is a normalized value. The actual reactance is −j 0.73 × 50 = −j 36.5 . It is close to the corresponding value of 36.6269 obtained earlier. Other components can be determined as well. Example 6.12 Two types of L-section matching networks are shown in Figure 6.28. Select one that can match the load ZL = 25 + j 10 to a 50- transmission line. Find the element values at 500 MHz. SOLUTION First, let us consider circuit (a). Rp must be 50 because it is required to be greater than Rs . However, the reactive element is connected in series with it. That will be possible only with Rs . Hence, this circuit cannot be designed using (6.3.8) to (6.3.10). The other set of design equations, namely, (6.3.11) to (6.3.13), requires admittance instead of impedance. Therefore, the given impedances are inverted to find the corresponding admittances as

224

IMPEDANCE-MATCHING NETWORKS

Ls

Cs

Z0 = 50 Ω

Lp

ZL

Z0 = 50 Ω

Cp

ZL

(b)

(a)

Figure 6.28 L-section matching circuits given for Example 6.12.

follows: YL =

1 1 = 0.034483 − j 0.01379 S = ZL 25 + j 10

Y0 =

1 1 = 0.02 S = Z0 50

and

Since Gs must be larger than Gp , this set of equations produces a matching circuit that has a reactance in series with YL . Thus, we conclude that the circuit given in Figure 6.28(a) cannot be designed. Now consider the circuit shown in Figure 6.28(b). For (6.3.8) to (6.3.10), Rp is 50 while Rs is 25 . Hence, one reactance of the matching circuit will be connected across 50 and the other one will go in series with ZL . The given circuit has this configuration. Hence, it will work. Its component values are calculated as 50 Q= −1=1 25 Therefore, Xs = Rs = 25 and Xp = Rp = 50 For Xp to be capacitive, we have 1 = 50 ⇒ Cp = 6.366 pF ωCp

225

MATCHING NETWORKS USING LUMPED ELEMENTS

As 10 of inductive reactance is already included in the load, another inductive 15 will suffice. Thus, ωLs = 25 − 10 = 15 ⇒ Ls =

15 H = 4.775 nH ω

This example can be solved using a ZY -Smith chart as well. To that end, the load impedance is normalized with characteristic impedance of the line. Hence, ZL =

25 + j 10 = 0.5 + j 0.2 50

This point is located on a ZY -Smith chart as shown in Figure 6.29. It is point A on this chart. If we move on the conductance circle from A, we never reach

90

100

110

80

70 60

0 12

50

0

15

30

0

14

40

0

13

10

160 170

20

B

A O 10

5

2

1

0.5

−20

0 −16

−10

−170

0.2

0

180

0

0 14

50 −1

−3 0

−

0

−4 −5

30

0

−6

0

20

−70

−80

−90

−100

−11 0

−1

−1

Figure 6.29 Graphical solution to Example 6.12.

226

IMPEDANCE-MATCHING NETWORKS

the unity-resistance circle. This means that an L-section matching circuit that has a reactance (inductor or capacitor) in parallel with the load cannot be designed. Therefore, circuit (a) of Figure 6.28 is not possible. Next, we try a reactance in series with the load. Moving on the constantresistance circle from A indicates that there are two possible circuits. Movement from A to B on the resistance circle and then from B to O on the unity conductance circle provides component values of circuit shown in Figure 6.28(b). The normalized reactance required is found to be j 0.3 (i.e., j 0.5 − j 0.2). It is an inductor because of its positive value. Further, it will be connected in series with the load because it is found by moving on a constant-resistance circle from A. The series inductance required is determined as XL = 0.3 × 50 = 15 ⇒ Ls =

15 H = 4.775 nH ω

Movement from B to O gives a normalized susceptance of j 1. Positive susceptance is a capacitor in parallel with a 50- line. Hence, ωCp =

1 ⇒ Cp = 6.366 pF 50

These results are found to be exactly equal to those found earlier analytically. SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P. A., Microwave Engineering, Englewood Cliffs, NJ: Prentice Hall, 1988. Sinnema, W., Electronic Transmission Technology. Englewood Cliffs, NJ: Prentice Hall, 1988.

PROBLEMS 6.1. A 10-V (rms) voltage source in series with a 50- resistance represents a signal generator. It is to be matched with a 100- load. Design a matching circuit that provides perfect matching over the frequency band 1 kHz to 1 GHz. Determine the power delivered to the load.

227

PROBLEMS

6.2. A 1000- load is to be matched with a signal generator that can be represented by a 1-A (rms) current source in parallel with a 100- resistance. Design a resistive circuit to match it. Determine the power dissipated in the matching network. 6.3. Design a single-stub network to match a 800 −j 300- load to a 400- lossless line. The stub should be located as close to the load as possible and it is to be connected in parallel with the transmission line. 6.4. A 140 −j 70- load is terminating a 70- transmission line, as shown in Figure P6.4. Find the location and length of a short-circuited stub of a 40- characteristic impedance that will match the load with the line.

70 Ω

70 Ω 140 − j70 Ω

40 Ω

S.C.

Figure P6.4

6.5. An antenna has impedance of 40 +j 30 at its input. Match it with a 50 line using short-circuited shunt stubs. Determine (a) the required stub admittance, (b) the distance between the stub and the antenna, (c) the stub length, and (d) the VSWR on each section of the circuit. 6.6. A lossless 100- transmission line is to be matched with a 100 +j 100 load using a double-stub tuner (Figure P6.6). Separation between the two stubs is λ/8 and the characteristic impedance is 100 . A load is connected right at the location of the first stub. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs.

228

IMPEDANCE-MATCHING NETWORKS

100 + j100 Ω 100 Ω

λ/8

Figure P6.6

6.7. A lossless 75- transmission line is to be matched with a 150 +j 15- load using a shunt-connected double-stub tuner. Separation between the two stubs is λ/8 and the characteristic impedance is 75 . The stub closest to the load (first stub) is λ/2 away from it. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs. 6.8. A lossless 50- transmission line is to be matched with a 25 +j 50- load using a double-stub tuner. Separation between the two stubs is 3λ/8 and the characteristic impedance is 50 . A load is connected at 0.2λ from the first stub, as shown in Figure P6.8. Determine the shortest possible lengths of the two stubs to obtain the matched condition, and find the VSWR between the two stubs.

0.2λ

3λ/8

Z0 = 50 Ω Z0 = 50 Ω

Z0 = 50 Ω

ZL = 25 + j50 Ω Z0 = 50 Ω

l2

l1

Figure P6.8

229

PROBLEMS

6.9. Complete or verify the two interstage designs of 1 GHz shown as (a) and (b) in Figure P6.9.

Γin = 0.29 ∠40 °

Matching network

ΓL = 0.58 ∠20 ° O.C.

~0.174λ

~0.086λ

(a)

(b)

Figure P6.9

6.10. Determine the shortest length d in the two networks in Figure P6.10 to match the load to a 50- lossless line. Also find the stub length l for circuit (a) and the required inductance L for circuit (b).

d

O.C. stub

d

ZL

ZL

L

ΓL = 0.3 ∠45° (a)

ΓL = 0.3 ∠45° (b)

Figure P6.10

230

IMPEDANCE-MATCHING NETWORKS

6.11. Determine the shortest length d in the two networks in Figure P6.11 to match a 100 −j 50- load to a 50- lossless line. Also find the required capacitance C for circuit (a) and the stub length l for circuit (b). d

d

50 Ω

C

ZL

50 Ω

ZL

l (a)

(b)

Figure P6.11

6.12. Complete or verify the two interstage designs of 1 GHz shown as (a) and (b) in Figure P6.12. Assume that the characteristic impedance is 50 .

Matching network

Γin = 1/3 ∠−120°

ΓL = 2/3 ∠50° O.C. C

Series stub

d

~ 0.1λ (b)

(a)

Figure P6.12

6.13. Complete or verify the interstage designs at f = 4 GHz, shown as (a) and (b) in Figure P6.13.

231

PROBLEMS

Matching network

Γin = 0.3 ∠45°

Load

ΓL = 0.3 ∠−90° d

d

Γin = 0.3 ∠45° L

Γin = 0.3 ∠45° C

ZL

ZL

(b)

(a)

Figure P6.13

6.14. Complete or verify the interstage designs at f = 4 GHz shown as (a) and (b) in Figure P6.14. The characteristic impedance is 50 .

Γin = 3/7 ∠80°

ΓL = 5/9∠−150° 0.126λ

~0.385λ

C Γin = 3/7 ∠80°

L ZL

Γin = 3/7 ∠ 80°

(a)

ZL

(b)

Figure P6.14

6.15. Two types of L-section matching networks are shown in Figure P6.15. Select one that can match the load ZL = 20 −j 100 to a 50- transmission line. Find the element values at 500 MHz.

232

IMPEDANCE-MATCHING NETWORKS C Ls

Lp

50 Ω

ZL

50 Ω

L

(a)

ZL

(b)

Figure P6.15

6.16. Two types of L-section matching networks are shown in Figure P6.16. Select one that can match a 30 +j 50- load to a 50- line at 1 GHz.

C1 C 50 Ω

L

ZL

ZL

50 Ω C2

(a)

(b)

Figure P6.16

6.17. Design a matching network (Figure P6.17) that will transform a 50 −j 50- load to the input impedance of 25 +j 25 .

Zin = 25 + j25 Ω

Matching network

ZL = 50 − j50 Ω

Figure P6.17

6.18. Match the load in Figure P6.18 to a 50- generator using lumped elements. Assume that the signal frequency as 4 GHz.

Γin = 0

Matching network

Load

ΓL = 0.5 ∠45°

Figure P6.18

233

PROBLEMS

6.19. Two types of L-section matching networks are shown in Figure P6.19. Select one that can match the load ZL = 60 −j 20 to a 50- transmission line. Find the element values at 500 MHz. Ls

50 Ω

Ls

Cp

ZL

50 Ω

(a)

Cp

ZL

(b)

Figure P6.19

6.20. Design a lumped-element network that will provide a load ZL = 30 +j 50 for the amplifier shown in Figure P6.20 operating at 2 GHz.

Lumped elements

Amplifier

50 Ω

ZL

Figure P6.20

6.21. Design a lumped-element network that will provide a load ZL = 30 +j 50 for the amplifier shown in Figure P6.21, and stop the biasing voltage from appearing across 50 . The amplifier is operating at 500 MHz.

Lumped elements

Amplifier

50 Ω

ZL

Figure P6.21

6.22. A 200- load is to be matched with a 50- line. Design (a) a resistive network and (b) a reactive network to match the load. (c) Compare the performance of the two designs.

7 IMPEDANCE TRANSFORMERS

In Chapter 6, several techniques were considered to match a given load impedance at a fixed frequency. These techniques included transmission line stubs as well as lumped elements. Note that lumped-element circuits may not be practical at higher frequencies. Further, it may be necessary in certain cases to keep the reflection coefficient below a specified value over a given frequency band. In this chapter we present transmission line impedance transformers that can meet such requirements. The chapter begins with a single-section impedance transformer that provides perfect matching at a single frequency. A matching bandwidth can be increased at the cost of a higher reflection coefficient. This concept is used to design multisection transformers. The characteristic impedance of each section is controlled to obtain the desired passband response. Multisection binomial transformers exhibit an almost flat reflection coefficient about the center frequency and increase gradually on either side. A wider bandwidth is achieved with an increased number of quarter-wave sections. Chebyshev transformers can provide even wider bandwidth with the same number of sections, but the reflection coefficient exhibits ripples in its passband. The chapter includes a procedure to design these multisection transformers as well as transmission line tapers. The chapter concludes with a brief discussion on the Bode–Fano constraints, which provide an insight into the trade-off between the bandwidth and the allowed reflection coefficient. 7.1 SINGLE-SECTION QUARTER-WAVE TRANSFORMERS We considered a single-section quarter-wavelength transformer design problem in Example 3.5. In this section we present a detailed analysis of such circuits. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

234

SINGLE-SECTION QUARTER-WAVE TRANSFORMERS

235

l

Z0 Z1

Zin

Figure 7.1

RL

Single-section quarter-wave transformer.

Consider the load resistance RL that is to be matched with a transmission line of characteristic impedance Z0 . Assume that a transmission line of length l and characteristic impedance Z1 is connected between the two, as shown in Figure 7.1. Its input impedance Zin is found as follows: Zin = Z1

RL + j Z1 tan βl Z1 + j RL tan βl

(7.1.1)

√ For βl = 90◦ (i.e., l = λ/4) and Z1 = Z0 RL , Zin is equal to Z0 ; hence, there is no reflected wave beyond this point toward the generator. However, it reappears at other frequencies when βl = 90◦ . The corresponding reflection coefficient in can be determined as follows: in = =

Zin − Z0 Z1 [(RL + j Z1 tan βl)/(Z1 + j RL tan βl)] − Z0 = Zin + Z0 Z1 [(RL + j Z1 tan βl)/(Z1 + j RL tan βl)] + Z0 RL − Z0 = ρin exp(j ϕ) √ RL + Z0 + j 2 Z0 RL tan βl

Therefore, ρin = =

[(RL + Z0

RL − Z0 + 4Z0 RL tan2 βl]1/2

)2

1 √ {1 + [2 Z0 RL /(RL − Z0 ) sec βl]2 }1/2

(7.1.2)

Variation in ρin with frequency is illustrated in Figure 7.2. For βl near 90◦ , it can be approximated as follows: |RL − Z0 | |RL − Z0 | ρin ≈ √ cos βl ≈ √ 2 Z0 RL tan βl 2 Z0 RL

(7.1.3)

236

IMPEDANCE TRANSFORMERS 0.35 0.30 0.25

ρin

0.20 0.15 0.10 0.05 0.00 0.0

ρM

0.5

θ1 1.0

π − θ1 1.5

2.0

2.5

3.0

3.5

βl

Figure 7.2 Reflection coefficient characteristics of a single-section impedance transformer used to match a 100- load to a 50- line.

If ρM is the maximum allowable reflection coefficient at the input, then √ 2ρM Z0 RL π θ1 < (7.1.4) cos θ1 = 2 (RL − Z0 ) 1 − ρ2M In the case of a TEM wave propagating on the transmission line, βl = (π/2) (f/f0 ), where f0 is the frequency at which βl = π/2. In this case, the bandwidth f2 − f1 = f is given by 2f0 (7.1.5) f = (f2 − f1 ) = 2(f0 − f1 ) = 2 f0 − θ1 π and the fractional bandwidth is

√ 2ρM Z0 RL f 4 −1 = 2 − cos f0 π 2 (RL − Z0 ) 1 − ρM

(7.1.6)

Example 7.1 Design a single-section quarter-wave impedance transformer to match a 100- load to a 50- air-filled coaxial line at 900 MHz (Figure 7.3). Determine the range of frequencies over which the reflection coefficient remains below 0.05.

237

MULTISECTION QUARTER-WAVE TRANSFORMERS

Z0

Z1

RL

Γin

Figure 7.3

Setup for Example 7.1.

SOLUTION For RL = 100 and Z0 = 50 , Z1 = and l=

√

100 × 50 = 70.7106781

λ 3 × 108 m = 8.33 cm = 4 4 × 900 × 106

The magnitude of the reflection coefficient increases as βl changes from π/2 (i.e., the signal frequency changes from 900 MHz). If the maximum allowed ρ is ρM = 0.05 (VSWR = 1.1053), the fractional bandwidth is found to be √ 2ρM Z0 RL 4 f −1 = 2 − cos = 0.180897 f0 π 2 (RL − Z0 ) 1 − ρM Therefore, 818.5964 MHz ≤ f ≤ 981.4037 MHz or 1.4287 ≤ βl ≤ 1.7129.

7.2 MULTISECTION QUARTER-WAVE TRANSFORMERS Consider an N -section impedance transformer connected between a transmission line of characteristic impedance of Z0 and load RL , as shown in Figure 7.4. As indicated, the length of every section is the same, although their characteristic impedances are different. Impedance at the input of N th section can be found as follows: exp(j βl) + N exp(−j βl) N Zin = ZN (7.2.1) exp(j βl) − N exp(−j βl) where N =

RL − ZN RL + ZN

(7.2.2)

The reflection coefficient seen by the (N − 1)st section is N−1 =

N Zin ZN (ej βl + N e−j βl ) − ZN−1 (ej βl − N e−j βl ) − ZN−1 = N ZN (ej βl + N e−j βl ) + ZN−1 (ej βl − N e−j βl ) Zin + ZN−1

238

IMPEDANCE TRANSFORMERS

x l l l ZN−1

ZN

RL

Z2 Z1 Z0

Figure 7.4 N-section impedance transformer.

or N−1 =

(ZN − ZN−1 )ej βl + N (ZN + ZN−1 )e−j βl (ZN + ZN−1 )ej βl + N (ZN − ZN−1 )e−j βl

Therefore, N−1 =

N−1 + N e−j 2βl 1 + N N−1 e−j 2βl

(7.2.3)

ZN − ZN−1 ZN + ZN−1

(7.2.4)

where N−1 =

If ZN is close to RL and ZN−1 is close to ZN , then N and N−1 are small quantities, and a first-order approximation can be assumed. Hence, N−1 ≈ N−1 + N e−j 2βl

(7.2.5)

Similarly, N−2 ≈ N−2 + N e−j 2βl = N−2 + N−1 e−j 2βl + N e−j 4βl

Therefore, by induction, the reflection coefficient seen by the feeding line is ≈ 0 + 1 e−j 2βl + 2 e−j 4βl + · · · + N−1 e−j 2(N−1)βl + N e−j 2Nβl or =

N

(7.2.6)

n e−j 2nβl

(7.2.7)

Zn+1 − Zn Zn+1 + Zn

(7.2.8)

n=0

where n =

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

239

Thus, we need a procedure to select n so that is minimized over the desired frequency range. To this end, we recast equation (7.2.6) as follows: = 0 + 1 w + 2 w2 + · · · + N wN = N

N

(w − wn )

(7.2.9)

n=1

where ϕ = −2βl

(7.2.10)

w = ej ϕ

(7.2.11)

and

Note that for βl = 0 (i.e., λ → ∞), individual transformer sections in effect have no electrical length and the load RL appears to be connected directly to the main line. Therefore, N RL − Z0 = n = (∵ w = 1) (7.2.12) RL + Z0 n=0 and only N of the (N + 1)-section reflection coefficients can be selected independently.

7.3 TRANSFORMER WITH UNIFORMLY DISTRIBUTED SECTION REFLECTION COEFFICIENTS If all of the section reflection coefficients are equal, (7.2.9) can be simplified as follows: wN+1 − 1 = 1 + w + w2 + w3 + · · · + wN−1 + wN = N w−1

(7.3.1)

or ej (N+1)ϕ − 1 sin{[(N + 1)/2]ϕ} = = ej Nϕ/2 j ϕ N e −1 sin(ϕ/2) Hence, sin{[(N + 1)/2]ϕ} = (N + 1)ρN sin{[(N + 1)/2]ϕ} || = ρ(ϕ) = ρN sin(ϕ/2) (N + 1) sin(ϕ/2) (7.3.2) and from (7.2.12), N RL − Z0 n = (N + 1)ρN = (7.3.3) RL + Z0 n=0

240

IMPEDANCE TRANSFORMERS

Therefore, equation (7.3.2) can be written as follows: RL − Z0 sin[(N + 1)βl] · ρ(βl) = RL + Z0 (N + 1) sin βl

(7.3.4)

This can be viewed as an equation that describes magnitude ρ of the reflection coefficient as a function of frequency. As (7.3.4) indicates, a pattern of ρ(βl) repeats periodically with an interval of π. It peaks at nπ, where n is an integer including zero. Further, there are N − 1 minor lobes between two consecutive main peaks. The number of zeros between the two main peaks of ρ(βl) is equal to the number of quarter-wave sections, N . Consider that there are three quarter-wave sections connected between a 100- load and a 50- line. Its reflection coefficient characteristics can be found from (7.3.4), as illustrated in Figure 7.5. There are three zeros in it, one at βl = π/2 and the other two located symmetrically around this point. In other words, zeros occur at βl = π/4, π/2, and 3π/4. When the number of quarterwave sections is increased from three to six, the ρ(βl) plot changes as illustrated in Figure 7.6. For a six-section transformer, Figure 7.6 shows five minor lobes between two main peaks of ρ(βl). One of these minor lobes has its maximum value (peak) at βl = π/2. Six zeros of this plot are located symmetrically: βl = nπ/7, n = 1, 2, . . . , 6. Thus, characteristics of ρ(βl) can be summarized as follows: ž ž

The pattern of ρ(βl) repeats with an interval of π. There are N nulls and N − 1 minor peaks in an interval. 0.35 0.30 0.25

ρ(βl)

0.20 0.15 0.10 0.05 0.00 0.0

P1 0.5

P2 1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.5 Reflection coefficient versus βl of a three-section transformer with equal section reflection coefficients for RL = 100 and Z0 = 50 .

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

241

0.35 0.30 0.25

ρ(βl)

0.20 0.15 0.10 0.05

P1

P2

0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.6 Reflection coefficient versus βl for a six-section transformer with equal section-reflection coefficients (RL = 100 and Z0 = 50 ). ž ž

When N is odd, one of the nulls occurs at βl = π/2 (i.e., l = λ/4). If ρM is specified as an upper bound on ρ to define the frequency band, points P1 and P2 bound the acceptable range of βl. This range becomes larger as N increases.

Since wN+1 − 1 =

N

(w − wn )

(7.3.5)

n = 1, 2, . . . , N

(7.3.6)

n=0

where wn = ej [2πn/(N+1)] (7.3.1) may be written as follows: N N = (w − wn ) = (w − ej [2πn/(N+1)] ) N n=1 n=1

(7.3.7)

This equation is of the form of (7.2.9). It indicates that when section reflection coefficients are the same, roots are equispaced around the unit circle on the complex w-plane with the root at w = 1 deleted. This is illustrated in Figure 7.7 for N = 3. It follows that N N ρ j [2πn/(N+1)]) = (w − e |(w − ej [2πn/(N+1)] )| (7.3.8) = ρN n=1

n=1

242

IMPEDANCE TRANSFORMERS Imaginary w

1 ϕ Real w

Figure 7.7

Location of zeros on a unit circle on the complex w-plane for N = 3. w3

d3

w2

d2

d1

w1

Figure 7.8

Graphical representation of (7.3.9) for N = 3 and ϕ = 0.

Since w = ej ϕ is constrained to lie on the unit circle, the distance between w and wn is given by dn (ϕ) = |ej ϕ − ej [2πn/(N+1)] | (7.3.9) For the case of N = 3, dn (ϕ = 0) is illustrated in Figure 7.8. From (7.3.3), (7.3.8), and (7.3.9), we get ρ(ϕ) =

N 1 RL − Z0 dn (ϕ) N + 1 RL + Z0 n=1

(7.3.10)

Thus, as ϕ = −2βl varies from 0 to 2π, w = ej ϕ makes one complete traverse of the unit circle, and distances d1 , d2 , . . . , dN vary with ϕ. If w coincides with the

TRANSFORMER WITH EQUAL SECTION REFLECTION COEFFICIENTS

243

root wn , the distance dn is zero. Consequently, the product of the distances is zero. Since the product of these distances is proportional to the reflection coefficient, ρ(ϕn ) goes to zero. It attains a local maximum whenever w is approximately halfway between successive roots. Example 7.2 Design a four-section quarter-wavelength impedance transformer with uniform distribution of reflection coefficient to match a 100- load to a 50- air-filled coaxial line at 900 MHz. Determine the range of frequencies over which the reflection coefficient remains below 0.1. Compare this bandwidth with that obtained for a single-section impedance transformer. SOLUTION From (7.3.3) with N = 4, we have 4 = and from (7.2.2),

1 1 100 − 50 = 5 100 + 50 15

1 100 − Z4 = 15 100 + Z4

or 100 + Z4 = 1500 − 15Z4 ⇒ 16Z4 = 1400 ⇒ Z4 = 87.5 The characteristic impedance of other sections can be determined from (7.2.8) as follows: 1 Z4 − Z3 = ⇒ 87.5 + Z3 = 15 × 87.5 − 15Z3 ⇒ Z3 = 76.5625 15 Z4 + Z3 Z3 − Z2 14 × 76.5625 1 = = 66.9922 ≈ 67 ⇒ Z2 = 15 Z3 + Z2 16 and Z2 − Z1 14 × 67 1 = = 58.625 ⇒ Z1 = 15 Z2 + Z1 16 The frequency range over which reflection coefficient remains below 0.1 is determined from (7.3.4) as follows: 1 sin 5θm sin 5θm = 1.5 0.1 = × ⇒ 3 5 sin θm sin θm where θm represents the value of βl at which magnitude of reflection coefficient is equal to 0.1. This is a transcendental equation that can be solved graphically. To that end, one needs to plot | sin 5θm | and 1.5 | sin θm | versus θm on the same graph and look for the intersection of two curves. Alternatively, a numerical

244

IMPEDANCE TRANSFORMERS 0.35 0.30 N=1

0.25 N=4 0.20

θ1

0.10

θ4

ρ

0.15

0.05 0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.9 Reflection coefficient versus βl for a four-section (N = 4) and a single section (N = 1) of impedance transformer.

method can be employed to determine θm . Two solutions to this equation are found to be 0.476 and 2.665, respectively. Hence, 0.476 ≤ βl ≤ 2.665 or

272.7 MHz ≤ f ≤ 1.5269 GHz

Corresponding bandwidth with a single section can be evaluated as follows: 1.2841 ≤ βl ≤ 1.8575 or

735.74 MHz ≤ f ≤ 1.0643 GHz

Reflection coefficient characteristics for two cases are displayed in Figure 7.9. Clearly, it has a much wider bandwidth, with four sections in comparison with that of a single section.

7.4 BINOMIAL TRANSFORMERS As shown in Figures 7.5 and 7.6, there are peaks and nulls in the passband of a multisection quarter-wavelength transformer with uniform section reflection coefficients. This characteristic can be traced to equispaced roots on the unit circle. One way to avoid this behavior is to place all the roots at a common pointw equal to −1. With this setting, distances dn are the same for all cases and N n=1 (w − wn ) goes to zero only once. It occurs for ϕ equal to −π (i.e., at βl = π/2). Thus, ρ is zero only for the frequency at which each section of

245

BINOMIAL TRANSFORMERS

the transformer is λ/4 long. With wn = −1 for all n, equation (7.2.9) may be written as N! wm = (w − wn ) = (w + 1)N = N m!(N − m)! n=1 m=0 N

N

(7.4.1)

The following binomial expansion is used in writing (7.4.1): (1 + x) = m

N n=1

m! xn n!(m − n)!

A comparison of equation (7.4.1) with (7.2.7) indicates that n N! = N n!(N − n)!

n = 0, 1, 2, . . . , N

(7.4.2)

and therefore the section reflection coefficients, normalized to N , are binomially distributed. From equation (7.4.1), = N (w + 1)N ⇒ (βl) = N (e−j 2βl + 1)N = N 2N e−j Nβl (cos βl)N or ρ(βl) = ρN 2N | cos βl|N

(7.4.3)

For βl = 0, load RL is effectively connected to the input line. Therefore, RL − Z0 ρ(0) = ρN 2 = RL + Z0 N

and

RL − Z0 × | cos βl|N ρ(βl) = RL + Z0

(7.4.4)

(7.4.5)

Reflection coefficient characteristics of multisection binomial transformers versus βl (in degrees) are illustrated in Figure 7.10. The reflection coefficient scale is normalized with the load reflection coefficient ρL . Unlike the preceding case of uniformly distributed section reflection coefficients, it shows a smooth characteristic without lobes. Example 7.3 Design a four-section quarter-wavelength binomial impedance transformer to match a 100- load to a 50- air-filled coaxial line at 900 MHz. Determine the range of frequencies over which the reflection coefficient remains below 0.1. Compare this result with that obtained in Example 7.2.

246

IMPEDANCE TRANSFORMERS 1.0 N=1 N = 20

Reflection coefficient

0.8 N=2

N=9

0.6 N=6 0.4

0.2

0.0

0

Figure 7.10

45

90 θ (deg)

135

180

Reflection coefficient versus βl for a multisection binomial transformer.

SOLUTION With N = 4 and n = 0, 1, and 2 in (7.4.2), we get 4! 0 4 = ⇒ 0 = 4 = 4 0!4! 4 3 4! 1 = = ⇒ 1 = 3 = 44 4 1!3! 4 and 4! 2 = = 6 ⇒ 2 = 64 4 2!2! From (7.4.4), RL − Z0 ⇒ ρ4 = 1 ρ(0) = ρN 2 = RL + Z0 24 N

100 − 50 1 100 + 50 = 48 = 0.020833

and from (7.2.8), n =

Zn+1 − Zn ⇒ ρn (Zn+1 + Zn ) = Zn+1 − Zn Zn+1 + Zn

Therefore, Zn =

1 − ρn × Zn+1 1 + ρn

(7.4.6)

1 + ρn × Zn 1 − ρn

(7.4.7)

Alternatively, Zn+1 =

BINOMIAL TRANSFORMERS

247

Characteristic impedance of each section can be determined from (7.4.6), as follows: Z4 =

1 − 1/48 × 100 = 95.9184 1 + 1/48

Z3 =

1 − 4/48 × 95.9184 = 81.1617 1 + 4/48

Z2 =

1 − 6/48 × 81.1617 = 63.1258 1 + 6/48

Z1 =

1 − 4/48 × 63.1258 = 53.4141 1 + 4/48

and

If we continue with this formula, we find that Z0 =

1 − 1/48 × 53.4141 = 51.2339 1 + 1/48

This is different from the given value of 50 . It happened because of the approximation involved in the formula. Error keeps building up if the characteristic impedances are determined proceeding in just one way. To minimize it, common practice is to determine the characteristic impedances up to halfway proceeding from the load side and then the remaining half from the input side. Thus, Z1 and Z2 should be determined from (7.4.7) as follows: Z1 =

1 + 1/48 × 50 = 52.1277 1 − 1/48

Z2 =

1 + 4/48 × 50 = 61.6054 1 − 4/48

and

The frequency range over which the reflection coefficient remains below 0.1 can be determined from (7.4.5) as follows: 0.1 = 13 | cos(ϑM )|4 ⇒ ϑM = 0.7376 Therefore, 0.7376 ≤ βl ≤ 2.404

or 422.61 MHz ≤ f ≤ 1.3774 GHz

Clearly, it has a larger frequency bandwidth than that of a single section. However, it is less than the bandwidth obtained with uniformly distributed section reflection coefficients. Reflection coefficient as a function of βl is illustrated in Figure 7.11.

248

IMPEDANCE TRANSFORMERS 0.35 0.30 0.25

ρ(βl)

0.20 θB

0.15 0.10 0.05 0.00 0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

βl

Figure 7.11

Reflection coefficient of a four-section binomial transformer versus βl.

7.5 CHEBYSHEV TRANSFORMERS Consider again the case of a uniform three-section impedance transformer that is connected between a 50- line and 100- load. Distribution of zeros around the unit circle is shown in Figure 7.12 with solid points. Its frequency response is illustrated in Figure 7.13 as curve (a). Now, let us move two of these zeros to ±120◦ while keeping the remaining one fixed at 180◦ , as illustrated by unfilled Imaginary w

Real w

Figure 7.12 Distribution of zeros for a uniform three-section impedance transformer (solid), with two of those zeros moved to ±120◦ (unfilled) or to ±60◦ (hatched).

249

CHEBYSHEV TRANSFORMERS 0.5

0.4 (a) 0.3

(b)

ρ(ϕ)

(c) 0.2

0.1

0 0

1

2

3

4

5

6

ϕ

Figure 7.13 Reflection coefficient versus ϕ for (a) a uniform three-section transformer, (b) two zeros moved to ±120◦ , and (c) two zeros moved to ±60◦ .

points in Figure 7.12. With this change in the distribution of zeros on the complex w-plane, ρ(βl) versus βl varies as shown by curve (b) in Figure 7.13. It shows relatively much lower peaks of intervening lobes, while the widths and heights of the two main lobes increase. On the other hand, if we move the two zeros to ±60◦ , the main peaks go down while intervening lobes rise. This is illustrated by hatched points in Figure 7.12 and by curve (c) in Figure 7.13. In this case, minor lobes increase while the main lobes are reduced in size. Note that zeros in this case are uniformly distributed around the unit circle, and therefore its ρ(βl) characteristic has identical lobes. Thus, the heights of intervening lobes decrease at the cost of the main lobe when zeros are moved closer together. On the other hand, moving the zeros farther apart raises the level of intervening lobes but reduces the main lobe. However, we need a systematic method to determine the location of each zero for a maximum permissible reflection coefficient, ρM , and the number of quarter-wave sections, N . An optimal distribution of zeros around the unit circle will keep the peaks of all passband lobes at the same height of ρM . In order to have the magnitudes of all minor lobes in the passband equal, section reflection coefficients are determined by the characteristics of Chebyshev functions, named after the Russian mathematician who first studied them. These functions satisfy the following differential equation: (1 − x 2 )

d 2 Tm (x) dTm (x) −x + m2 Tm (x) = 0 2 dx dx

(7.5.1)

Chebyshev functions of degree m, represented by Tm (x), are mth-degree polynomials that satisfy (7.5.1). The first four of these and a recurrence relation for

250

IMPEDANCE TRANSFORMERS

higher-order Chebyshev polynomials are given as follows: T1 (x) = x T2 (x) = 2x 2 − 1 T3 (x) = 4x 3 − 3x T4 (x) = 8x 4 − 8x 2 + 1 .. . Tm (x) = 2xTm−1 (x) − Tm−2 (x) Alternatively, −1 cos(m cos x) − 1 ≤ x ≤ 1 Tm (x) = cosh(m cosh−1 x) x ≥ 1 (−1)m cosh(m cosh−1 |x|) x ≤ −1 Note that for x = cos θ,

(7.5.2)

Tm (cos θ) = cos mθ

(7.5.3)

Figure 7.14 depicts Chebyshev polynomials of degrees 1 through 4. The following characteristics can be noted: ž ž

The magnitudes of these polynomials oscillate between ±1 for −1 ≤ x ≤ 1. For |x| > 1, |Tm (x)| increases at a faster rate with x as m increases.

6.0 5.0 4.0 3.0 m=4

2.0

m=3

m=2

m=1

1.0 0.0 −1.0 −2.0 −3.0 −1.5

−x0 −1.0

x0 −0.5

0.0

0.5

1.0

Figure 7.14 Chebyshev polynomials for m = 1, 2, 3, 4.

1.5

251

CHEBYSHEV TRANSFORMERS ž

ž

The numbers of zeros are equal to the order of the polynomials. Zeros of an even-order polynomial are located symmetrically about the origin, with one of the minor lobes’ peak at x = 0. Polynomials of odd orders have one zero at x = 0 while the remaining zeros are located symmetrically.

These characteristics of Chebyshev polynomials are utilized to design an impedance transformer that has ripples of equal magnitude in its passband. The number of quarter-wave sections determines the order of Chebyshev polynomials and the distribution of zeros on the complex w-plane is selected according to that. With x0 properly selected, |Tm (x)| corresponds precisely to ρ(βl). This is done by linking βl to x of Chebyshev polynomial as follows: x = x0 cos βl

(7.5.4)

Consider the design of a three-section equal-ripple impedance transformer. A Chebyshev polynomial of order 3 is appropriate for this case. Figure 7.15 illustrates T3 (x) together with (7.5.4). It is to be noted that Chebyshev variable x and angle ϕ on the complex w-plane are related through (7.5.4) because ϕ is equal to −2βl. As ϕ varies from 0 to −2π, x changes as illustrated in Figure 7.15(b) and the corresponding T3 (x) in Figure 7.15(a). Figure 7.16 shows |T3 (ϕ)|, which can represent the desired ρ(ϕ) provided that (RL − Z0 ) (R + Z ) T3 (ϕ = 0) T3 (x0 ) L 0 = = 1 1 ρM

(7.5.5)

where ρM is the maximum allowed reflection coefficient in the passband. For an m-section impedance transformer, (7.5.5) can be written as RL − Z0 1 Tm (x0 ) = RL + Z0 ρM

(7.5.6)

Location x0 can now be determined from (7.5.2) as follows:

1 −1 x0 = cosh × cosh Tm (x0 ) m

(7.5.7)

With x0 determined from (7.5.7), |Tm (x)| represents the ρ(βl) desired. Hence, zeros of the reflection coefficient are the same as those of Tm (x). Since Chebyshev polynomials have zeros in the range −1 < x < 1, zeros of ρ(ϕ) can be determined from (7.5.2). Therefore,

π Tm (xn ) = 0 ⇒ cos(m × cos−1 xn ) = 0 = ± cos (2n − 1) 2

n = 1, 2, . . .

252

IMPEDANCE TRANSFORMERS 2.0

(a)

0.0

−2.0 (b)

x0 cos βl

0

βl π

−1.5

−1.0

−0.5

0.0

0.5

1.0

1.5

Figure 7.15 Third-order Chebyshev polynomial (a) and its variable, x versus βl (b). 2.5

T3(x0)

RL − Z0 2.0 RL + Z 0 1.5

ρM 1.0

0.5

0.0 −7.0

Figure 7.16

−6.0

−5.0

−4.0

−3.0

−2.0

−1.0

0.0

|T3 (x)| versus ϕ and its relationship with the desired ρ(ϕ).

where xn is the location of the nth zero. Hence,

π xn = ± cos (2n − 1) 2m

(7.5.8)

and the corresponding ϕn can be evaluated from (7.5.4) as follows: ϕn = 2 cos−1

xn x0

(7.5.9)

253

CHEBYSHEV TRANSFORMERS

The zeros of ρ(ϕ), wn , on the complex w-plane are now known because wn = eiϕn . Equation (7.2.9) can be used to determine the section reflection coefficient, n . Zn is, in turn, determined from equation (7.2.8). The bandwidth of the impedance transformer extends from x = −1 to x = +1. With βl = θM at x = 1, (7.5.4) gives 1 cos θM = x0 Therefore, bandwidth of a Chebyshev transformer can be expressed as follows: cos−1

1 1 ≤ βl ≤ π − cos−1 x0 x0

(7.5.10)

Example 7.4 Reconsider the matching of a 100- load to a 50- line. Design an equal-ripple four-section quarter-wavelength impedance transformer and compare its bandwidth with those obtained earlier for ρM = 0.1. SOLUTION From (7.5.6), 100 − 50 × 1 = 1 = 3.3333 T4 (x0 ) = 100 + 50 0.1 0.3 Therefore, x0 can now be determined from (7.5.7) as follows: x0 = cosh 14 cosh−1 (3.3333) In case inverse hyperbolic functions are not available on a calculator, the following procedure can be used to evaluate x0 . Assume that y = cosh−1 (3.3333) ⇒ 3.3333 = cosh y = (ey + e−y )/2. Hence, ey + e−y = 2 × 3.3333 = 6.6666 or e

2y

− 6.6666 × e + 1 = 0 ⇒ e = y

y

6.6666 ±

(6.6666)2 − 4 = 6.5131 2

and 0.1535y = ln(6.5131) = 1.8738 Therefore,

1.8738 1 cosh−1 (3.3333) = = 0.4684 4 4

and x0 = cosh(0.4684) =

e0.4684 + e−0.4684 = 1.1117 2

Note that the other solution of ey produces y = −1.874 and x0 = 1.1117.

254

IMPEDANCE TRANSFORMERS

From (7.5.8) and (7.5.9), we get xn = ±0.9239, ±0.3827 and

◦

◦

◦

◦

ϕn = −67.61 , −139.71 , −220.29 , −292.39 Therefore, w1 = 0.381 − j 0.925 w2 = −0.763 − j 0.647 w3 = −0.763 + j 0.647 and w4 = 0.381 + j 0.925 Now, from equation (7.2.9) we get

= 4 (1 + 0.764w + 0.837w2 + 0.764w3 + w4 ) Therefore, 0 = 4 , 1 = 3 = 0.7644

and 2 = 0.8374

From equation (7.2.12), we have =

N

n =

RL − Z0 RL − Z0 ⇒ 4.3654 = RL + Z0 RL + Z0

4 =

100 − 50 1 × = 0.076 4.365 100 + 50

n=0

Hence, 0 = 4 = 0.076,

1 = 3 = 0.058,

and 2 = 0.064

Now, the characteristic impedance of each section can be determined from (7.2.8) as follows: ZL − Z4 1 − 4 ⇒ Z4 = ZL = 85.87 ZL + Z4 1 + 4 Z4 − Z3 1 − 3 ⇒ Z3 = Z4 = 76.46 3 = Z4 + Z3 1 + 3 Z3 − Z2 1 − 2 2 = ⇒ Z2 = Z3 = 67.26 Z3 + Z2 1 + 2

4 =

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

255

and 1 =

Z2 − Z1 1 − 1 ⇒ Z1 = Z2 = 59.89 Z2 + Z1 1 + 1

To minimize the accumulating error in Zn , it is advisable to calculate half of the impedance values from the load side and the other half from the input side. In other words, Z1 and Z2 should be determined as follows: Z1 =

1 + 0 Z0 = 58.2251 1 − 0

Z2 =

1 + 1 Z1 = 65.3951 1 − 1

and

The bandwidth is determined from (7.5.10) as follows: cos βl =

1 = 0.899 ⇒ βl = 0.452 x0

Therefore, 0.452 ≤ βl ≤ 2.6896 Thus, the bandwidth is greater than what was achieved from either a uniform or a binomial distribution of n coefficients.

7.6 EXACT FORMULATION AND DESIGN OF MULTISECTION IMPEDANCE TRANSFORMERS The analysis and design presented so far is based on the assumption that the section reflection coefficients are small, as implied by (7.2.5). If this is not the case, an exact expression for the reflection coefficient must be used. Alternatively, there are design tables available in the literature (e.g., Matthaei et al., 1980) that can be used to synthesize an impedance transformer. In case of a two- or threesection Chebyshev transformer, the design procedure summarized below may be used as well. Exact formulation of the multisection impedance transformer is conveniently developed via the power loss ratio, PLR , defined as follows: PLR =

incident power power delivered to the load

If Pin represents the incident power and ρin is the input reflection coefficient, then Pin 1 PLR − 1 = ⇒ ρin = (7.6.1) PLR = PLR (1 − ρ2in )Pin 1 − ρ2in

256

IMPEDANCE TRANSFORMERS

For any transformer, ρin can be determined from its input impedance Zin . The power loss ratio PLR can subsequently be evaluated from (7.6.1). This PLR can be expressed in terms of Q2N (cos θ), an even polynomial of degree 2N in cos θ. Hence, PLR = 1 + Q2N (cos θ) (7.6.2) Coefficients of Q2N (cos θ) are functions of various impedances Zn . For an equalripple characteristic in the passband, a Chebyshev polynomial can be used to specify PLR as follows: PLR = 1 + k 2 TN2 (sec θM cos θ)

(7.6.3)

where k 2 is determined from the maximum value of PLR in the passband. θM represents the value of βl that corresponds to the maximum allowed reflection coefficient ρM . Since TN2 (sec θM cos θ) has a maximum value of unity in the passband, the extreme value of PLR will be 1 + k 2 . Various characteristic impedances are determined by equating (7.6.2) and (7.6.3). Further, (7.6.1) produces a relation between ρM and k 2 , as follows: k2 ρM = (7.6.4) 1 + k2 For the two-section impedance transformer shown in Figure 7.17, reflection coefficient characteristics as a function of θ are illustrated in Figure 7.18. The power loss ratio for this transformer is found to be PLR = 1 +

(ZL − Z0 )2 (sec2 θz cos2 θ − 1)2 4ZL Z0 tan4 θz

(7.6.5)

where θz is the value of θ at the lower zero (i.e., θz < π/2), as shown in Figure 7.18. The maximum power loss ratio, PLR max , is found to be PLR

max

Z0

=1+

Z1

(ZL − Z0 )2 cot4 θz 4ZL Z0

Z2

(7.6.6)

ZL

θ θ

Figure 7.17

Two-section impedance transformer.

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

257

|Γin |

ρM

θ θM

θZ

π/2

π − θM

Figure 7.18 Reflection coefficient characteristics of a two-section Chebyshev transformer.

Hence, k2 =

(ZL − Z0 )2 cot4 θz 4ZL Z0

(7.6.7)

Required values of characteristic impedances, Z1 and Z2 , are determined from the equations Z12 = Z02

ZL (ZL − Z0 )2 + 2 4 Z0 4Z0 tan θz

and Z2 =

1/2 +

(ZL − Z0 )Z0 2 tan2 θz

ZL Z0 Z1

(7.6.8)

(7.6.9)

The passband edge, θM , is given by √ θM = cos−1 ( 2 cos θz )

(7.6.10)

Hence, the bandwidth of this transformer is found as θM ≤ θ = βl ≤ π − θM

(7.6.11a)

258

IMPEDANCE TRANSFORMERS

For a TEM wave,

4θM f =2− f0 π

(7.6.11b)

If the bandwidth is specified in a design problem along with ZL and Z0 , then θM is known. Therefore, θz can be determined from (7.6.10). Impedances Z1 and Z2 are determined subsequently from (7.6.8) and (7.6.9), respectively. The corresponding maximum reflection coefficient ρM can easily be evaluated following k 2 from (7.6.7). On the other hand, if ρM is specified instead of θM , then k 2 is determined from (7.6.4). θz , in turn, is calculated from (7.6.7). Now, Z1 , Z2 , and θM can be determined from (7.6.8), (7.6.9), and (7.6.10), respectively. In the limit θz → π/2, two zeros of ρ in Figure 7.18 come together to give a maximally flat transformer. In that case, (7.6.8) and (7.6.9) are simplified to give Z1 = ZL × Z0

1/4

3/4

(7.6.12)

3/4 ZL

1/4 Z0

(7.6.13)

Z2 =

×

and θM = cos−1 (cot θz )

(7.6.14)

Example 7.5 Use the exact theory to design a two-section Chebyshev transformer to match a 500- load to a 100- line. The required fractional bandwidth is 0.6. What is the resulting value of ρM ? SOLUTION From (7.6.10) and (7.6.11), √ 2 − f/f0 4 1 f −1 = 2 − cos ( 2 cos θz ) ⇒ cos θz = √ cos f0 π 4/π 2 2 − 0.6 1 = 0.321 = √ cos 4/π 2 Therefore, θz = 1.244 rad and √ √ θM = cos−1 ( 2 cos θz ) = cos−1 ( 2 × 0.321) = 1.1 From (7.6.7), k2 =

(ZL − Z0 )2 (Z L − 1)2 cot4 θz = cot4 θz = 0.0106 4ZL Z0 4Z L

Therefore, from (7.6.4), ρM =

k2 = 0.1022 1 + k2

259

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

Now, from (7.6.8),

1/2 ZL (ZL − Z0 )Z0 (ZL − Z0 )2 2 + + ⇒ Z1 Z0 2 tan2 θz 4Z02 tan4 θz 1/2 (Z L − 1)2 ZL − 1 = + ZL + = 2.4776 4 tan4 θz 2 tan2 θz

Z12 = Z02

Therefore, Z1 =

√

2.4776 = 1.57 ⇒ Z1 = 157.41

and from (7.6.9), Z2 =

ZL Z0 = 317.65 Z1

Check : For θ = π/2, 2

Z in =

Z1

Z = 2 L

Z2

=

Z in − 1 Z in + 1

1.57412 (5) = 1.2277 3.17652 = 0.1022 = ρm

Let us consider the design of a three-section impedance transformer. Figure 7.19 shows a three-section impedance transformer connected between the load ZL and a transmission line of characteristic impedance Z0 . The reflection coefficient characteristic of such a Chebyshev transformer is illustrated in Figure 7.20. The power loss ratio, PLR , of a three-section Chebyshev transformer is found to be PLR = 1 +

Z0

(ZL − Z0 )2 (sec2 θz cos2 θ − 1)2 cos2 θ 4ZL Z0 tan4 θz

(7.6.15)

ZL

Z1 Z2

Z3

θ θ θ

Figure 7.19 Three-section impedance transformer.

260

IMPEDANCE TRANSFORMERS

| Γin |

ρM θ θM θZ

Figure 7.20 former.

Reflection coefficient characteristics of a three-section Chebyshev trans-

It attains the maximum allowed value at θ = θM , where 2 θM = cos−1 √ cos θz 3

(7.6.16)

Since the maximum power loss ratio must be equal to 1 + k 2 , k2 =

(ZL − Z0 )2 4ZL Z0

2 cos θz √ 3 3 tan2 θz

2 (7.6.17)

Characteristic impedance Z1 is obtained by solving the equation 1/2 1/2 ZL Z02 Z12 ZL ZL ZL − Z0 = +2 Z1 − −2 Z1−1 Z02 tan2 θz Z0 Z0 Z0 Z12

(7.6.18)

The other two characteristic impedances, Z2 and Z3 , are then determined as follows: Z2 = ZL Z0 (7.6.19) and Z3 =

ZL Z0 Z1

(7.6.20)

EXACT ANALYSIS OF MULTISECTION TRANSFORMERS

261

The range of the passband (bandwidth) is still given by (7.6.11) provided that θM is now computed from (7.6.16). Example 7.6 Design a three-section Chebyshev transformer (exact theory) to match a 500- load to a 100- line. The required fractional bandwidth is 0.6. Compute ρM and compare it with that obtained in Example 7.5. SOLUTION From (7.6.16) and (7.6.11), (π/2) − θM 2 f = 0.6 ⇒ and √ cos θz f0 π/2 3 π = 0.3 ⇒ θM = (1 − 0.3) = 1.1 2

θM = cos−1

Therefore,

√ cos θz =

3 cos θM = 0.3932 ⇒ θz = 1.17 rad 2

Now, from (7.6.17), (ZL − Z0 )2 k = 4ZL Z0 2

2 cos θz √ 3 3 tan2 θz

2

and from (7.6.4),

(ZL − 1)2 = 4ZL

2 cos θz √ 3 3 tan2 θz

2

= 6.125 × 10−4

ρM =

k2 = 0.0247 1 + k2

which is approximately one-fourth of the previous (two-section transformer) case. Z1 is obtained by solving equation (7.6.18) as follows: 1/2 1/2 ZL Z02 ZL − Z0 Z12 ZL ZL = + 2 Z − − 2 Z1−1 Z02 1 tan2 θz Z0 Z0 Z0 Z12 or 4 ZL − 1 ZL 2 −1 = Z 1 + 2(Z L )1/2 Z 1 − 2 − 2(Z L )1/2 Z 1 = = 0.7314 2θ tan2 θz tan z Z1 Therefore, 2 Z1

or

√ √ 5 2 5 + 2 5 Z1 − 2 − = 0.7314 Z1 Z1

√ √ 4 3 2 Z 1 + 2 5 Z 1 − 0.7314 Z 1 − 2 5 Z 1 − 5 = 0

262

IMPEDANCE TRANSFORMERS

Therefore, Z 1 = −4.4679, (−0.6392 − j 0.6854), (−0.6392 + j 0.6854), and 1.2741 Thus, only one of these solutions can be physically realized because the others have a negative real part. After selecting Z1 = 127.42 , (7.6.19) and (7.6.20) give Z3 =

ZL Z0 = 392.4277 Z1

Z2 =

ZL Z0 = 223.6068

and

If we use the approximate formulation, then following the procedure of Example 7.4, we find from (7.5.6) that T3 (x0 ) =

500 − 100 1 × = 26.9458 500 + 100 0.02474

Now, from (7.5.7), x0 = cosh[ 31 cosh−1 (26.9458)] = 2.0208 and from (7.5.8),

π = ±0.866, 0 xn = ± cos (2n − 1) 6 Using (7.5.9) we get ϕn = 2.2558, 3.1416, and 4.0274 Since wn = ej ϕn , we find that w1 = −0.6327 + j 0.7744 w2 = −0.6327 − j 0.7744 w3 = −1 and from (7.2.9), = (w − w1 )(w − w2 )(w − w3 ) = w3 + 2.2654w2 + 2.2654w + 1 3 Hence, 0 = 3 1 = 2 = 2.26543

TAPERED TRANSMISSION LINES

263

Now, from (7.2.12), 0 + 1 + 2 + 3 = 6.53073 = Therefore,

500 − 100 500 + 100

3 = 0.1021 = 0 1 = 2 = 0.2313

Corresponding impedances are now determined from (7.2.8) as follows: 1 + 0 Z1 = = 1.2274 1 − 0 calculated from the input side 1 + 1 Z2 = Z 1 = 2.4234 1 − 1 and 1 − 3 Z3 = Z L = 4.0737 calculated from the load side 1 + 3 A comparison of these values with those obtained earlier using the exact formulation shows that the two sets are within 10% of deviation for this example. Note that for the fractional bandwidth of 0.6 with a single section, reflection coefficient ρM increases to 0.3762. 7.7 TAPERED TRANSMISSION LINES Consider that the multisection impedance transformer of Figure 7.4 is replaced by a tapered transition of length L, as illustrated in Figure 7.21. The characteristic impedance of this transition is a continuous smooth function of distance, with its values at the two ends as Z0 and RL . An approximate theory of such a matching section can easily be developed on the basis of the analysis presented in Section 7.2. The continuously tapered line can be modeled by a large number of incremental sections of length δz. One of these sections, connected at z, has a characteristic impedance of Z + δZ, and the one before it has a characteristic impedance of Z, as shown in Figure 7.21. These impedance values are conveniently normalized by Z0 before obtaining the incremental reflection coefficient at distance z. Hence, δ0 =

Z + δZ − Z Z + δZ + Z

≈

δZ 2Z

(7.7.1)

As δz → 0, it can be written as d0 =

dZ 2Z

=

1 d(ln Z) dz 2 dz

(7.7.2)

264

IMPEDANCE TRANSFORMERS

L

δz

Z Z0

RL

Z + δZ

z

Figure 7.21 Tapered transition connected between load RL and transmission line of characteristic impedance Z0 .

The corresponding incremental reflection coefficient din at the input can be written as follows: d −j 2βz −j 2βz 1 d0 = e (ln Z) dz din ≈ e 2 dz The total reflection coefficient, in , at the input of the tapered section can be determined by summing up these incremental reflections with their appropriate phase angles. Hence,

L

in = 0

din =

1 2

L

e−j 2βz

0

d (ln Z) dz dz

(7.7.3)

Therefore, in can be determined from (7.7.3) provided that Z(z) is given. However, the synthesis problem is a bit complex because in that case, Z(z) is to be determined for a specified in . Let us first consider a few examples of evaluating in for the given distributions of d(ln Z)/dz. Case 1

d (ln Z) is constant over the entire length of the taper. Suppose that dz d ln Z(z) = C1 dz

0≤z≤L

where C1 is a constant. On integrating this equation, we get ln Z(z) = C1 z + C2 With Z(z = 0) = 1 and Z(z = L) = R L , constants C1 and C2 can be determined. Hence, z ln Z = ln R L ⇒ Z = e(z/L) ln RL (7.7.4) L

265

TAPERED TRANSMISSION LINES

Thus, the impedance is changing exponentially with distance. Therefore, this kind of taper is called an exponential taper. From (7.7.3), the total reflection coefficient is found as L ln R L L −j 2βz 1 L −j 2βz d z ln R L e−j 2βz in = e e dz = ln R L dz = 2 0 dz L 2L 0 2L −j 2β 0 or in =

ln R L e−j βL − 1 sin βL ln R L −j βL sin βL 2ρin = = e ⇒ 2L −j 2β 2 βL βL ln R L

(7.7.5)

where ρin = |in |. It is assumed in this evaluation that β = 2π/λ is not changing with distance z. The right-hand side of (7.7.5) is displayed in Figure 7.22 as a function of βL. Since L is fixed for a given taper and β is related directly to signal frequency, this plot also represents the frequency response of an exponential taper. d d (ln Z) is a triangular function. If (ln Z) is a triangular function, dz dz defined as 4z L ln R L 0≤z≤ 2 d(ln Z) L2 = (7.7.6) 1 L z dz 4 ln R L − ≤z≤L L L2 2 Case 2

1.0

0.8

0.6

0.4

0.2

0.0 0.0000

3.1416

6.2832

9.4248 βL

12.5664

15.7080

18.8496

Figure 7.22 Frequency response of an exponential taper.

21.9912

266

then

or

IMPEDANCE TRANSFORMERS

L/2 4z ln R L dz 0 L2 ln Z = 1 z L ln R L dz − L/2 4 L L2 2 2z 2 ln R L + c1 L ln Z = z z2 4 ln R L + c2 − L 2L2

0≤z≤

L 2

L ≤z≤L 2 0≤z≤

L 2

L ≤z≤L 2

Integration constants c1 and c2 are determined from the given conditions Z(z = 0) = 1 and Z(z = L) = R L . Hence, 2 L 2z 0≤z≤ exp L2 ln R L 2 (7.7.7) Z= 2 4z 2z L − ≤ z ≤ L exp − 1 ln R L L L2 2 and therefore total reflection coefficient is found from (7.7.3) as in =

1 2

L/2 0

or in =

e−j 2βz

2 ln R L L2

L/2

4z ln R L dz + L2

ze−j 2βz dz +

0

L L/2

L

e−j 2βz

4 (L − z) ln R L dz L2

(L − z)e−j 2βz dz

L/2

Since

and

ze−j 2βz dz =

e−j 2βz dz =

ze−j 2βz e−j 2βz − −j 2β (−j 2β)2 e−j 2βz −j 2β

in can be found as follows: in =

1 −j βL sin(βL/2) 2 ln R L e 2 βL/2

(7.7.8)

Figure 7.23 shows normalized magnitude of in as a function of βL. As before, it also represents the frequency response of this taper of length L. When we compare it with that of Figure 7.22 for an exponential taper, we find that the

267

TAPERED TRANSMISSION LINES 1.0

0.8

0.6

0.4

0.2

0.0 0.0000

3.1416

6.2832

9.4248

12.5664

15.7080

18.8496

21.9912

βL

Figure 7.23 Frequency response of a taper with triangular distribution.

magnitude of minor lobes is reduced now, but the main lobe has become relatively wider. It affects the cutoff frequency of a given taper. For example, if the maximum allowed normalized reflection coefficient is 0.2, an exponential taper has a lower value of βL than that of a taper with triangular distribution. However, it is the other way around for a normalized reflection coefficient of 0.05. For a fixed length L, βL is proportional to frequency, and hence the lower value of βL has a lower cutoff frequency than in the other case. Figure 7.24 depicts variation in normalized characteristic impedance along the length of two tapers. A coaxial line operating in the TEM mode can be used to design these tapers by varying the diameter of its inner conductor. Similarly, the narrow sidewall width of a TE10 -mode rectangular waveguide can be modified to get the desired taper while keeping its wide side constant. Case 3

d (ln Z) is a Gaussian distribution. Assume that dz d 2 (ln Z) = K1 e−α(z−L/2) dz

(7.7.9)

This distribution is centered around the midpoint of the tapered section. Its fall-off rate is governed by the coefficient α. Integrating it over distance z, we find that

z

ln Z = K1 0

e−α(z −L/2) dz 2

(7.7.10)

268

IMPEDANCE TRANSFORMERS 2.0

1.8 (b)

(a)

1.6

1.4

1.2

1.0 0.0

0.2

0.4

0.6

0.8

1.0

Figure 7.24 Distribution of normalized characteristic impedance along a tapered line terminated by RL = 2Z0 for (a) case 1 and (b) case 2.

Since Z(z = L) = R L , ln R L

K1 =

L

e

−α(z −L/2)2

(7.7.11) dz

0

Therefore,

z

e−α(z −L/2) dz

ln Z = ln R L 0L

2

(7.7.12) e

−α(z −L/2)2

dz

0

and in =

1 K1 2

L

e−j 2βz e−α(z−L/2) dz 2

(7.7.13)

0

With the following substitution of variable and associated limits of integration in (7.7.11) and (7.7.13), z−

L =x 2 dz = dx z=0→x=−

L 2

269

TAPERED TRANSMISSION LINES

and z=L→x=

L 2

we get in =

1 K1 2

L/2

−L/2

1 K1 e−j βL 2

e−j 2β(x+L/2) e−αx dx =

or

2

1 in = ln R L e−j βl 2

L/2

L/2

e−j 2βx e−αx dx

−L/2

2

L/2

e

−αx 2

2

e−j 2βx e−αx dx

1 = ln R L e−j βl 2

−L/2

dx

L/2

0

cos 2βx e−αx dx 2

−L/2

L/2

e−αx dx 2

0

(7.7.14)

It can be arranged after substituting

√

α x with y, and

1 in = ln R L e−j βl 2

ξ 0

√

α(L/2) with ξ as follows:

cos[(βL/ξ)y]e−y dy ξ 2 e−y dy 2

(7.7.15)

0

Figure 7.25 shows the normalized magnitude of the reflection coefficient versus βL for two different values of ξ. It indicates that the main lobe becomes wider, and the level of minor lobes goes down as ξ is increased.

1 0.8 (a) 0.6 0.4

(b)

0.2 0 0

5

10

15

20

βL

Figure 7.25 Frequency response of a tapered line with a normal distribution of normalized characteristic impedance for (a) ξ = 0.3 and (b) ξ = 3.

270

IMPEDANCE TRANSFORMERS

7.8 SYNTHESIS OF TRANSMISSION LINE TAPERS The frequency response of a transmission line taper can be determined easily from (7.7.3). However, the inverse problem of determining its impedance variation that provides the desired frequency response needs further consideration. To this end, assume that F (α) is the Fourier transform of a function f (x). In other words, ∞ f (x)e−j αx dx (7.8.1) F (α) = −∞

and

1 f (x) = 2π

∞

F (α)ej αx dα

(7.8.2)

−∞

For convenience, we rewrite (7.7.3) as follows: L 1 d in (2β) = (ln Z) e−j 2βz dz 2 dz 0 If

1 2

d(ln Z)/dz = 0 for −∞ ≤ z < 0 and for z > L, it may be written as ∞ 1 d (7.8.3) in (2β) = (ln Z)e−j 2βz dz −∞ 2 dz

A comparison of (7.8.3) with (7.8.1) indicates that in (2β) represents the Fourier transform of 12 d[ln Z(z)]/dz. Hence, from (7.8.2), ∞ 1 1 d(ln Z) = in (2β)ej 2βx d(2β) (7.8.4) 2 dz 2π −∞ This equation can be used to design a taper that will have the desired reflection coefficient characteristics. However, only those in (2β) that have an inverse Fourier transform limited to 0 ≤ z ≤ L (zero outside this range) can be realized. At this point, we can conveniently introduce the following normalized variables: p = 2π

z − (L/2) L

(7.8.5)

and u=

2L βL = π λ

Therefore, z=

L (p + π) 2π

and dz =

L dp 2π

(7.8.6)

SYNTHESIS OF TRANSMISSION LINE TAPERS

New integration limits are found as −π ≤ p ≤ π. Hence, 1 π −j 2β(L/2+Lp/2π) d dp L (ln Z) dp in (2β) = e 2 −π dp dz 2π or π 1 d in (2β) = e−j βL e−j up (ln Z) dp 2 dp −π Now, if we define g(p) = d(ln Z)/dp and F (u) = can be expressed as follows:

π

−π

271

(7.8.7)

e−j up g(p) dp, (7.8.7)

in (2β) = 12 e−j βL F (u) or ρin (2β) = 12 |F (u)| Further, F (u = 0) = ln Z L

(7.8.8)

Thus, F (u) and g(p) form the Fourier transform pair. Therefore, ∞ 1 nonzero for |p| < π jup e F (u) du = g(p) = 0 for |p| > π 2π −∞

(7.8.9)

To satisfy the conditions embedded in writing (7.8.3), only those F (u) that produce g(p) as zero for |p| > π can be synthesized. Suitable restrictions on F (u) can be derived after expanding g(p) in a complex Fourier series as follows: ∞ a ejnp −π ≤ p ≤ π n g(p) = n=−∞ (7.8.10) 0 |p| > π ∗ Since g(p) is a real function, constants an = a−n . Therefore,

F (u) =

π

e

−j up

−π

∞

an e

n=−∞

jnp

dp =

∞

an

n=−∞

π

e −π

−j (u−n)p

π e−j (u−n)p dp = an −j (u − n) −π n=−∞ ∞

or F (u) =

∞ n=−∞

an

∞ e−j (u−n)π − ej (u−n)π sin π(u − n) = 2π an −j (u − n) π(u − n) n=−∞

For u = m (an integer), sin π(m − n) = π(m − n)

1 for m = n 0 for m = n

(7.8.11)

272

IMPEDANCE TRANSFORMERS

Therefore, F (n) = 2πan ⇒ an =

F (n) 2π

and F (u) =

∞

F (n)

n=−∞

sin π(u − n) π(u − n)

(7.8.12)

This is the well-known sampling theorem used in communication theory. It states that F (u) is uniquely reconstructed from a knowledge of sample values of F (u) at u = n, where n is an integer (positive or negative) including zero (i.e., u = n = 0, ±1, ±2, ±3, . . .). To have greater flexibility in selecting F (u), let us assume that an = 0 for all |n| > N . Therefore, N an ejnp (7.8.13) g(p) = n=−N

and from (7.8.11), F (u) = 2π

N n=−N

an (−1)n

N sin πu u sin πu an (−1)n = 2π π(u − n) πu n=−N u−n

This series can be recognized as the partial-fraction expansion of a function, and therefore F (u) can be expressed as follows: F (u) = 2π

sin πu Q(u) N 2 2 πu n=1 (u − n )

(7.8.14)

where Q(u) is an arbitrary polynomial of degree 2N in u subject to the restriction ∗ . Further, it contains an arbitrary constant mulQ(−u) = Q∗ (u), so that an = a−n tiplier. In (7.8.14), sin πu has zeros at u = ±n. However, the first 2N of these zeros are canceled by (u2 − n2 ). These can be replaced by 2N new arbitrarily located zeros by proper choice of Q(u). Example 7.7 Design a transmission line taper to match a 100- load to a 50 line. The desired frequency response has a triple zero at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution. SOLUTION The desired characteristic can be accomplished by moving zeros at ±1 and ±3 into the points u = ±2. Due to this triple zero, the reflection coefficient will remain small over a relatively wide range of βL around ±2π. Therefore, Q(u) = C(u2 − 4)3

SYNTHESIS OF TRANSMISSION LINE TAPERS

and F (u) = 2πC

273

sin πu (u2 − 4)3 πu (u2 − 1) × (u2 − 4) × (u2 − 9)

Since u = βL/π, F (u) versus u represents the frequency response of this taper. Further, it can be normalized with F (0) as follows: F (u) (u2 − 4)3 = 9 sin πu F (0) 16 πu (u2 − 1)(u2 − 4)(u2 − 9) A plot of this equation is illustrated in Figure 7.26. It shows that the normalized magnitude is very small around u = 2. Since F (0) is given by (7.8.8), multiplying constant C in F (u) can be determined easily. Hence, F (0) = 2πC × 1 ×

(−4)3 9 16 = 2πC × = ln Z L ⇒ C = ln Z L (−1)(−4)(−9) 9 32π

Further, 2πC F (1) = π

(1 − 4)3 sin πu 9 × = Cπ u2 − 1 u=1 (1 − 4)(1 − 9) 8

9 9 ln Z L π = 0.3164 ln Z L 32π 8 F (2) = 0 25 (9 − 4)3 2πC sin πu = − Cπ × F (3) = 3π u2 − 9 u=3 (9 − 1)(9 − 4) 72 25 9 = − π × ln Z L = 0.09766 ln Z L 72 32π =

and F (n) = 0 for n > 3. Therefore, 1 1 F (0) = ln Z L 2π 2π 1 0.3164 a1 = a−1 = F (1) = ln Z L 2π 2π a2 = a−2 = 0

a0 =

a3 = a−3 =

0.09766 1 F (3) = − ln Z L 2π 2π

and an = a−n = 0

for n > 3

274

IMPEDANCE TRANSFORMERS 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

u

Figure 7.26 Normalized frequency response of the taper in Example 7.7.

Hence, g(p) =

d(ln Z) ln Z L = (a0 + 2a1 cos p + 2a3 cos 3p) dp 2π

=

ln Z L (1 + 0.6328 × cos p − 0.1953 × cos 3p) 2π

and ln Z =

ln Z L (p + 0.6328 × sin p − 0.0651 × sin 3p) + c1 2π

The integration constant c1 can be evaluated as follows: ln Z = 0 at p = −π or

ln Z = ln Z L at p = π

Therefore, c1 = 0.5 ln Z L and ln Z = where

ln Z L (p + π + 0.6328 sin p − 0.0651 sin 3p) 2π L 2π z− p= L 2

Normalized impedance (Z) versus the length of the taper (z/L) is illustrated in Figure 7.27.

275

SYNTHESIS OF TRANSMISSION LINE TAPERS 2.0

Normalized impedance

1.8 1.6 1.4 1.2 1.0 0.8 0.0

0.2

0.4

0.6

0.8

1.0

z/L

Figure 7.27 Normalized characteristic impedance variation along the length of the taper in Example 7.7.

Example 7.8 Design a taper to match a 100- load to a 50- transmission line. Its reflection coefficient has double zeros at βL = ±2π and ±3π. Plot its frequency response and normalized characteristic impedance versus the normalized length of the taper. SOLUTION This design can be achieved by moving the zeros at u = ± 1 and ± 4 into the points ± 2 and ± 3, respectively. Hence, Q(u) = C(u2 − 22 )2 (u2 − 32 )2 = C(u2 − 4)2 (u2 − 9)2 and F (u) = C

sin πu (u2 − 4)2 (u2 − 9)2 πu (u2 − 1)(u2 − 4)(u2 − 9)(u2 − 16)

Therefore, F (0) = 2.25C, F (1) = 0.8C, F (2) = 0, F (3) = 0, F (4) = (7/40)C, and F (n) = 0 for n > 4. From (7.8.8), F (0) = ln Z L ⇒ C = and g(p) =

∞ d(ln Z) an e−jnp = dp n=−∞

ln Z L 2.25 an =

F (n) 2π

or g(p) = =

2.25C 7C 0.8C jp + (e + e−jp ) + (ej 4p + e−j 4p ) 2π 2π (40)(2π) ln Z L (1 + 0.7111 cos p + 0.1556 cos 4p) 2π

276

IMPEDANCE TRANSFORMERS 1 0.8 0.6 0.4 0.2 0 0

2

4

6

8

u

Figure 7.28 Frequency response of the taper in Example 7.8.

Therefore, ln Z =

ln Z L (p + 0.7111 sin p + 0.0389 sin 4p) + c1 2π

Since, ln Z|p=−π = 0, c1 = ln Z L /2 and ln Z =

ln Z L (p + π + 0.7111 sin p + 0.0389 sin 4p) 2π

where p=

2π L z− L 2

Frequency response and normalized characteristic impedance distribution are shown in Figures 7.28 and 7.29, respectively.

Klopfenstein Taper The preceding procedure indicates infinite possibilities of synthesizing a transmission line taper. A designer naturally looks for the best design. In other words, which design gives the shortest taper for a given ρM ? The answer to this question is the Klopfenstein taper, which is derived from a stepped Chebyshev transformer. The design procedure is summarized here without its derivation. The characteristic impedance variation of this taper versus distance z is given by Z(z) =

Z0 RL eκ

(7.8.15)

277

SYNTHESIS OF TRANSMISSION LINE TAPERS 2.0

Normalized impedance

1.8

1.6

1.4

1.2

1.0

0.8 0.0

0.2

0.4

0.6

0.8

1.0

z /L

Figure 7.29 Normalized characteristic impedance variation of the taper in Example 7.8.

where

A2 0 2z f − 1, A cosh A L ζ I1 (A 1 − y 2 ) dy f (ζ, A) = A 1 − y2 0 κ=

0 =

0≤z≤L |x| ≤ 1

RL − Z0 RL + Z0

(7.8.16) (7.8.17) (7.8.18)

Since the cutoff value of βL is equal to A, we have A = β0 L

(7.8.19)

For a maximum ripple ρM in its passband, it can be determined as A = cosh−1

|0 | ρM

(7.8.20)

I1 (x) is the modified Bessel function of the first kind. f (ζ, A) needs to be evaluated numerically except for the following special cases: f (0, A) = 0 x f (x, 0) = 2 cosh A − 1 f (1, A) = A2

278

IMPEDANCE TRANSFORMERS

The resulting input reflection coefficient as a function of βL (and hence, frequency) is given by in (βL) = 0 e

−j βL cosh

A2 − (βL)2 cosh A

(7.8.21)

Note that the hyperbolic cosine becomes the cosine function for an imaginary argument. Example 7.9 Design a Klopfenstein taper to match a 100- load to a 50- transmission line. The maximum allowed reflection coefficient in its passband is 0.1. Plot the characteristic impedance variation along its normalized length and the input reflection coefficient versus βL. SOLUTION From (7.8.18), 0 =

RL − Z0 100 − 50 1 = = RL + Z0 100 + 50 3

TABLE 7.1 Characteristic Impedance of the Klopfenstein Taper (in Example 7.9) as a Function of Its Normalized Length Z/L

Impedance ()

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

56.000896 57.030043 58.169255 59.412337 60.757724 62.202518 63.74238 65.371433 67.082194 68.865533 70.710678 72.605261 74.535428 76.486009 78.440749 80.382597 82.294063 84.157604 85.95606 87.673089 89.29361

279

SYNTHESIS OF TRANSMISSION LINE TAPERS 90.0

Characteristic impedance (Ω)

85.0 80.0 75.0 70.0 65.0 60.0 55.0 0.0

0.2

0.4

0.6

0.8

1.0

z/L

Figure 7.30 Characteristic impedance distribution along the normalized length of the Klopfenstein taper in Example 7.9.

0.3 0.25

|Γin(βL)|

0.2 0.15 0.1 0.05 0 0

5

10

15

20

βL

Figure 7.31 Reflection coefficient versus βL of the Klopfenstein taper in Example 7.9.

From (7.8.20), A = cosh−1

1 |0 | = cosh−1 3 = 1.87382 ρM 0.1

Equation (7.8.15) is used to evaluate Z(z) numerically. These results are shown in Table 7.1. These data are also displayed in Figure 7.30. The reflection coefficient of this taper is depicted in Figure 7.31.

280

IMPEDANCE TRANSFORMERS

7.9 BODE–FANO CONSTRAINTS FOR LOSSLESS MATCHING NETWORKS Various matching circuits described in this and Chapter 6 indicate that zero reflection is possible only at selected discrete frequencies. Examples 6.6 and 6.7 show that the bandwidth over which the reflection coefficient remains below a specified value is increased with the number of components. A circuit designer would like to know whether a lossless passive network could be designed for perfect matching. Further, it will be helpful if associated constraints and design trade-offs are known. Bode–Fano constraints provide such means to the designer. Bode–Fano criteria provide optimum results under ideal conditions. These results may require approximation to implement the circuit. Table 7.2 summarizes these constraints for selected R –C and R –L loads. The detailed formulations are beyond the scope of this book. As an example, if it is desired to design a reactive matching network at the output of a transistor amplifier, the situation is similar to that depicted in the top row of the table. Further, consider the case in which magnitude ρ of the reflection TABLE 7.2 Bode–Fano Constraints to Match Certain Loads Using Passive Lossless Network Circuit Arrangement

Constraint Relation

Γ = ρ∠θ

Matching network

0

R

Γ = ρ∠θ

∞

C

Matching network

L

R

∞

0

π 1 ln dω ≤ ρ RC

ω2 πL 1 ln dω ≤ 0 ρ R

(where ω0 is the center frequency) Γ = ρ∠θ

Matching network

∞

C R

0

1 ln dω ≤ ω20 πRC ρ

(where ω0 is the center frequency)

Γ = ρ∠θ

Matching network

L

R 0

∞

πR 1 ln dω ≤ ρ L

PROBLEMS

281

coefficient remains constant at ρm (ρm < 1) over the frequency band from ω1 to ω2 , whereas it stays at unity outside this band. The constraint for this circuit is found to be ∞ ω2 1 1 1 π (7.9.1) ln dω = ln dω = (ω2 − ω1 ) ln ≤ ρ ρ ρ RC m ω1 m 0 This condition shows the trade-off associated with the bandwidth and the reflection coefficient. If R and C are given, the right-hand side of (7.9.1) is fixed. In this situation, a wider bandwidth (ω2 − ω1 ) is possible only at the cost of a higher reflection coefficient ρm . Further, ρm cannot go to zero unless the bandwidth ω2 − ω1 is zero. In other words, a perfect matching condition is achievable only at discrete frequencies.

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures. Dedham, MA: Artech House, 1980. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998.

PROBLEMS 7.1. The output impedance of a microwave receiver is 50 . Find the required length and characteristic impedance of a transmission line that will match a 100- load to the receiver. Signal frequency and phase velocity are 10 GHz and 2.4 × 108 m/s, respectively. What is the frequency band over which the reflection coefficient remains below 0.1? 7.2. A quarter-wave impedance-matching transformer is used to match a microwave source with internal impedance 10 to a 50- transmission line. If the source frequency is 3 GHz and the phase velocity on the line is equal to the speed of light in free space, find (a) the required length and characteristic impedance of the matching section and (b) the reflection coefficients at two ends of the matching section. 7.3. A 0.7-m-long transmission line short-circuited at one end has an input impedance of −j 68.8 . The signal frequency and phase velocity are 100 MHz and 2 × 108 m/s, respectively.

282

IMPEDANCE TRANSFORMERS

(a) What is the characteristic impedance of the transmission line? (b) If a 200- load is connected to the end of this line, what is the new input impedance? (c) A quarter-wave matching transformer is to be used to match a 200- load with the line. What will be the required length and characteristic impedance of the matching transformer? 7.4. Design a two-section binomial transformer to match a 100- load to a 75 line. What is the frequency band over which the reflection coefficient remains below 0.1? 7.5. Design a three-section binomial transformer to match a 75- load to a 50- transmission line. What is the fractional bandwidth for a VSWR less than 1.1? 7.6. Using the approximate theory, design a two-section Chebyshev transformer to match a 100- load to a 75- line. What is the frequency band over which the reflection coefficient remains below 0.1? Compare the results with those obtained in Problem 7.4. 7.7. Design a four-section Chebyshev transformer to match a network with input impedance 60 to a transmission line of characteristic impedance 40 . Find the bandwidth if the maximum VSWR allowed in its passband is 1.2. 7.8. Using the exact theory, design a two-section Chebyshev transformer to match a 10- load to a 75- line. Find the bandwidth over which the reflection coefficient remains below 0.05. 7.9. Using the exact theory, design a three-section Chebyshev transformer to match a 10- load to a 75- line. Find the bandwidth over which the reflection coefficient remains below 0.05. Compare the results with those obtained in Problem 7.8. 7.10. Design a transmission line taper to match a 75- load to a 50- line. The desired frequency response has a double zero at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution. 7.11. Design a transmission line taper to match a 40- load to a 75- line. Its reflection coefficient has triple zeros at βL = ±2π. Plot its frequency response and normalized characteristic impedance distribution versus the normalized length of the taper. 7.12. Design a Klopfenstein taper to match a 40- load to a 75- line. The maximum allowed reflection coefficient in its passband is 0.1. Plot the characteristic impedance variation along its normalized length and the input reflection coefficient versus βL. Compare the results with those obtained in Problem 7.11.

8 TWO-PORT NETWORKS

Electronic circuits are frequently needed for processing a given electrical signal to extract the desired information or characteristics. This includes boosting the strength of a weak signal or filtering out certain frequency bands, and so on. Most of these circuits can be modeled as a black box that contains a linear network comprising resistors, inductors, capacitors, and dependent sources. Thus, it may include electronic devices but not the independent sources. Further, it has four terminals, two for input and the other two for output of the signal. There may be a few more terminals to supply the bias voltage for electronic devices. However, these bias conditions are embedded in equivalent dependent sources. Hence, a large class of electronic circuits can be modeled as two-port networks. Parameters of the two-port completely describe its behavior in terms of voltage and current at each port. These parameters simplify the description of its operation when the two-port network is connected into a larger system. Figure 8.1 shows a two-port network along with appropriate voltages and currents at its terminals. Sometimes, port 1 is called the input port and port 2, the output port. The upper terminal is customarily assumed to be positive with respect to the lower one on either side. Further, currents enter the positive terminals at each port. Since the linear network does not contain independent sources, the same currents leave respective negative terminals. There are several ways to characterize this network. Some of these parameters and relations among them are presented in this chapter, including impedance parameters, admittance parameters, hybrid

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

283

284

TWO-PORT NETWORKS I1

I2 +

+ V1

Port 1

Linear network

Port 2

V2 −

−

Figure 8.1

Two-port network.

parameters, and transmission parameters. Scattering parameters are introduced later in the chapter to characterize the high-frequency and microwave circuits. 8.1 IMPEDANCE PARAMETERS Consider the two-port network shown in Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 at port 1 can be expressed in terms of two currents as follows: V1 = Z11 I1 + Z12 I2 (8.1.1) Since V1 is in volts and I1 and I2 are in amperes, parameters Z11 and Z12 must be in ohms. Therefore, these are called the impedance parameters. Similarly, we can write V2 in terms of I1 and I2 as follows: V2 = Z21 I1 + Z22 I2 Using the matrix representation, we can write Z11 I1 V1 Z12 = V2 Z21 Z22 I2

(8.1.2)

(8.1.3)

or [V ] = [Z][I ]

(8.1.4)

where [Z] is called the impedance matrix of the two-port network. If port 2 of this network is left open, I2 will be zero. In this condition, (8.1.1) and (8.1.2) give V1 Z11 = (8.1.5) I1 I2 =0 and Z21

V2 = I1 I2 =0

(8.1.6)

Similarly, with a source connected at port 2 while port 1 is open circuit, we find that V1 Z12 = (8.1.7) I2 I1 =0

285

IMPEDANCE PARAMETERS

and Z22 =

V2 I2 I1 =0

(8.1.8)

Equations (8.1.5) through (8.1.8) define the impedance parameters of a twoport network. Example 8.1 Find the impedance parameters for the two-port network shown in Figure 8.2. SOLUTION If I2 is zero, V1 and V2 can be found from Ohm’s law as 6I1 . Hence, from (8.1.5) and (8.1.6), V1 6I1 Z11 = = = 6 I1 I2 =0 I1 and Z21 =

V2 6I1 = = 6 I1 I2 =0 I1

Similarly, when a source is connected at port 2 and port 1 has an open circuit, we find that V2 = V1 = 6I2 Hence, from (8.1.7) and (8.1.8), Z12 = and Z22 Therefore,

V1 6I2 = = 6 I2 I1 =0 I2

V2 6I2 = = = 6 I2 I1 =0 I2

Z11 Z21

Z12 Z22

6 = 6

6 6

I1

I2 + V1 −

Figure 8.2

+ 6Ω

V2 −

Two-port network for Example 8.1.

286

TWO-PORT NETWORKS I1

+

+

V1

I2

V2

12 Ω

3Ω −

−

Figure 8.3

Two-port network for Example 8.2.

Example 8.2 Find the impedance parameters of the two-port network shown in Figure 8.3. SOLUTION As before, assume that a source is connected at port 1 while port 2 is open. In this condition, V1 = 12I1 and V2 = 0. Therefore, V1 12I1 Z11 = = = 12 I1 I2 =0 I1 and Z21

V2 = =0 I1 I2 =0

Similarly, when a source connected at port 2 and port 1 has an open circuit, we find that V2 = 3I2 and V1 = 0 Hence, from (8.1.7) and (8.1.8), Z12 and Z22 Therefore,

Z11 Z21

V1 = =0 I2 I1 =0

V2 3I2 = = = 3 I2 I1 =0 I2 Z12 Z22

12 = 0

0 3

Example 8.3 Find the impedance parameters for the two-port network shown in Figure 8.4. SOLUTION Assuming that the source is connected at port 1 while port 2 is open, we find that V1 = (12 + 6)I1 = 18I1

and V2 = 6I1

287

IMPEDANCE PARAMETERS 12 Ω I1

3Ω

+

+

V1

I2

V2

6Ω

−

−

Figure 8.4

Two-port network for Example 8.3.

Note that there is no current flowing through a 3- resistor because port 2 is open. Therefore, V1 18I1 = = 18 Z11 = I1 I2 =0 I1 and Z21

V2 6I1 = = = 6 I1 I2 =0 I1

Similarly, with a source at port 2 and port 1 open circuit, V2 = (6 + 3)I2 = 9I2

and V1 = 6I2

This time, there is no current flowing through a 12- resistor because port 1 is open. Hence, from (8.1.7) and (8.1.8), V1 6I2 Z12 = = = 6 I2 I1 =0 I2 and Z22 Therefore,

Z11 Z21

V2 9I2 = = = 9 I2 I1 =0 I2 Z12 Z22

=

18 6

6 9

An analysis of results obtained in Examples 8.1 to 8.3 indicates that Z12 and Z21 are equal for all three circuits. In fact, it is an inherent characteristic of these networks. It will hold for any reciprocal circuit. If a given circuit is symmetrical, Z11 will be equal to Z22 as well. Further, impedance parameters obtained in Example 8.3 are equal to the sum of the corresponding results found in Examples 8.1 and 8.2. This happens because if the circuits of these two examples are connected in series, we end up with the circuit of Example 8.3, shown in Figure 8.5.

288

TWO-PORT NETWORKS I1

I2 +

+ V1

3Ω 12 Ω

−

V2 −

I1

I2 +

+ 6Ω

V ′1

V ′2 −

−

Figure 8.5 Circuit for Example 8.3.

Example 8.4 Find the impedance parameters for a transmission line network shown in Figure 8.6. SOLUTION This circuit is symmetrical because interchanging port 1 and port 2 does not affect it. Therefore, Z22 must be equal to Z11 . Further, if current I at port 1 produces an open-circuit voltage V at port 2, current I injected at port 2 will produce V at port 1. Hence, it is a reciprocal circuit. Therefore, Z12 will be equal to Z21 . Assume that the source is connected at port 1 while the other port is open. If Vin is incident voltage at port 1, Vin e−γl is the voltage at port 2. Since the reflection coefficient of an open circuit is +1, the reflected voltage at this port is equal to the incident voltage. Therefore, the reflected voltage reaching port 1 is Vin e−2γl . Hence, V1 = Vin + Vin e−2γl V2 = 2Vin e−γl Vin I1 = (1 − e2γl ) Z0 l I1

+

V1

−

+ Z0, γ

V2

−

Figure 8.6 Transmission line network for Example 8.4.

I2

ADMITTANCE PARAMETERS

289

and I2 = 0 Therefore, Z11 =

V1 Vin (1 + e−2γl ) Z0 e+γl + e−γl = Z0 +γl = Z0 coth γl = = −2γl I1 I2 =0 (Vin /Z0 )(1 − e ) e − e−γl tanh γl

and Z21

V2 2Vin e−γl Z0 2 = = = = Z0 +γl −2γl −γl I1 I2 =0 (Vin /Z0 )(1 − e ) e −e sinh γl

For a lossless line, γ = j β, and therefore, Z11 =

Z0 = −j Z0 cot βl j tan βl

and Z21 =

Z0 Z0 = −j j sin βl sin βl

8.2 ADMITTANCE PARAMETERS Consider again the two-port network shown in Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, current I1 at port 1 can be expressed in terms of two voltages: I1 = Y11 V1 + Y12 V2 (8.2.1) Since I1 is in amperes and V1 and V2 are in volts, parameters Y11 and Y12 must be in siemens. Therefore, these are called the admittance parameters. Similarly, we can write I2 in terms of V1 and V2 as follows: I2 = Y21 V1 + Y22 V2 Using the matrix representation, we can write I1 Y11 Y12 V1 = I2 Y21 Y22 V2 or [I ] = [Y ][V ] where [Y ] is called the admittance matrix of the two-port network.

(8.2.2)

(8.2.3)

(8.2.4)

290

TWO-PORT NETWORKS

If port 2 of this network has a short circuit, V2 will be zero. In this condition, (8.2.1) and (8.2.2) give I1 (8.2.5) Y11 = V1 V2 =0 and Y21 =

I2 V1 V2 =0

(8.2.6)

Similarly, with a source connected at port 2 and a short circuit at port 1, I1 (8.2.7) Y12 = V2 V1 =0 and Y22

I2 = V2 V1 =0

(8.2.8)

Equations (8.2.5) through (8.2.8) define the admittance parameters of a twoport network. Example 8.5 Find the admittance parameters of the circuit shown in Figure 8.7. SOLUTION If V2 is zero, I1 is equal to 0.05 V1 and I2 is −0.05 V1 . Hence, from (8.2.5) and (8.2.6), I1 0.05 V1 Y11 = = = 0.05 S V1 V2 =0 V1 and Y21

I2 −0.05 V1 = = = −0.05 S V1 V2 =0 V1

Similarly, with a source connected at port 2 and port 1 having a short circuit, I2 = −I1 = 0.05 V2

0.05 S I1

+ V1 −

+ V2 −

Figure 8.7 Circuit for Example 8.5.

I2

291

ADMITTANCE PARAMETERS

Hence, from (8.2.7) and (8.2.8), Y12

I1 −0.05 V2 = = = −0.05 S V2 V1 =0 V2

Y22

I2 0.05 V2 = = = 0.05 S V2 V1 =0 V2

and

Therefore,

Y11 Y21

Y12 Y22

0.05 −0.05 = −0.05 0.05

Again we find that Y11 is equal to Y22 because this circuit is symmetrical. Similarly, Y12 is equal to Y21 because it is reciprocal. Example 8.6 Find the admittance parameters for the two-port network shown in Figure 8.8. SOLUTION Assuming that a source is connected at port 1 and port 2 has a short circuit, we find that I1 =

0.1(0.2 + 0.025) 0.0225 V1 = V1 0.1 + 0.2 + 0.025 0.325

A

and if voltage across 0.2 S is VN , then VN =

0.0225 I1 V1 = V1 = 0.2 + 0.025 0.225 × 0.325 3.25

V

Therefore, I2 = −0.2 VN = −

0.2 V1 3.25

0.1 S I1

0.2 S

+ V1 −

Figure 8.8

A

+

0.025 S

V2 −

Two-port network for Example 8.6.

I2

292

TWO-PORT NETWORKS

Hence, from (8.2.5) and (8.2.6), I1 0.0225 Y11 = = = 0.0692 S V1 V2 =0 0.325 and Y21 =

I2 0.2 = −0.0615 S =− V1 V2 =0 3.25

Similarly, with a source at port 2 and port 1 having a short circuit, I2 =

0.2(0.1 + 0.025) 0.025 V1 = V2 0.2 + 0.1 + 0.025 0.325

A

and if voltage across 0.1 S is VM , then VM =

I2 2 V2 0.025 = V2 = 0.1 + 0.025 0.125 × 0.325 3.25

Therefore, I1 = −0.1 VM = −

0.2 V2 3.25

V

A

Hence, from (8.2.7) and (8.2.8), I1 0.2 Y12 = =− = −0.0615 S V2 V1 =0 3.25 and Y22 Therefore,

I2 0.025 = = = 0.0769 S V2 V1 =0 0.325

Y11 Y21

Y12 Y22

=

0.0692 −0.0615

−0.0615 0.0769

As expected, Y12 = Y21 but Y11 = Y22 . This is because the given circuit is reciprocal but is not symmetrical. Example 8.7 Find the admittance parameters of the two-port network shown in Figure 8.9. SOLUTION Assuming that a source is connected at port 1 and port 2 has a short circuit, we find that 0.1(0.2 + 0.025) I1 = 0.05 + A V1 = 0.1192 V1 0.1 + 0.2 + 0.025

293

ADMITTANCE PARAMETERS 0.05 S 0.1 S I1

0.2 S +

+ V1

0.025 S

I2

V2 −

−

Figure 8.9

Two-port network for Example 8.7.

and if current through 0.05 S is IN , then IN =

0.05 I1 = 0.05 V1 0.05 + [0.1(0.2 + 0.025)/(0.1 + 0.2 + 0.025)]

A

Current through 0.1 S is I1 − IN = 0.0692 V1 . Using the current-division rule, current IM through 0.2 S is found as follows: IM =

0.2 0.0692 V1 = 0.0615 V1 0.2 + 0.025

A

Hence, I2 = −(IN + IM ) = −0.1115 V1 A. Now, from (8.2.5) and (8.2.6), Y11

I1 = = 0.1192 S V1 V2 =0

and Y21 =

I2 = −0.1115 S V1 V2 =0

Similarly, with a source at port 2 and port 1 having a short circuit, current I2 at port 2 is

0.2(0.1 + 0.025) I2 = 0.05 + V2 = 0.1269 V2 0.2 + 0.1 + 0.025

A

and current IN through 0.05 S can be found as follows: IN =

0.05 I2 = 0.05 V2 0.05 + [0.2(0.1 + 0.025)/(0.2 + 0.1 + 0.025)]

A

294

TWO-PORT NETWORKS 0.05 S + V1 −

I1

+ V2 −

I2

+

+

V1 −

V2

0.1 S

0.2 S

− +

+ V1

V2

0.025 S

−

Figure 8.10

−

Circuit for Example 8.7.

Current through 0.2 S is I2 − IN = 0.0769 V2 . Using the current-division rule one more time, the current IM through 0.1 S is found as follows: IM =

0.1 0.0769 V2 = 0.0615V2 0.1 + 0.025

A

Hence, I1 = −(IN + IM ) = −0.1115 V2 A. Therefore, from (8.2.7) and (8.2.8), Y12

I1 = = −0.1115 S V2 V1 =0

Y22

I2 = = 0.1269 S V2 V1 =0

and

Therefore,

Y11 Y21

Y12 Y22

=

0.1192 −0.1115

−0.1115 0.1269

As expected, Y12 = Y21 but Y11 = Y22 . This is because the given circuit is reciprocal but is not symmetrical. Further, we find that the admittance parameters obtained in Example 8.7 are equal to the sum of the corresponding impedance parameters of Examples 8.5 and 8.6. This is because when the circuits of these two examples are connected in parallel, we end up with the circuit of Example 8.7, shown in Figure 8.10. Example 8.8 Find the admittance parameters of a transmission line of length l, shown in Figure 8.11.

295

ADMITTANCE PARAMETERS l I1

+

+ Z0, γ

V1

−

I2

V2

−

Figure 8.11 Setup for Example 8.8.

SOLUTION This circuit is symmetrical because interchanging port 1 and port 2 does not affect it. Therefore, Y22 must be equal Y11 . Further, if voltage V at port 1 produces a short-circuit current I at port 2, voltage V at port 2 will produce current I at port 1. Hence, it is a reciprocal circuit. Therefore, Y12 will be equal to Y21 . Assume that a source is connected at port 1 when the other port has a short circuit. If Vin is the incident voltage at port 1, it will appear as Vin e−γl at port 2. Since the reflection coefficient of a short circuit is equal to −1, reflected voltage at this port is 180◦ out of phase with incident voltage. Therefore, the reflected voltage reaching at port 1 is −Vin e−2γl . Hence, V1 = Vin − Vin e−2γl V2 = 0 Vin (1 + e−2γl ) I1 = Z0 and I2 = −

2Vin −γl e Z0

Therefore, Y11

e+γl + e−γl 1 I1 (Vin /Z0 )(1 + e−2γl ) = = = = −2γl +γl −γl V1 V2 =0 Vin (1 − e ) Z0 (e − e ) Z0 tanh γl

Y21

I2 −(2Vin /Z0 )e−γl 2 1 = = =− =− −2γl +γl −γl V1 V2 =0 Vin (1 − e ) Z0 (e − e ) Z0 sinh γl

and

For a lossless line, γ = j β and therefore Y11 =

1 j Z0 tan βl

and Y21 = −

1 1 =j j Z0 sin βl Z0 sin βl

296

TWO-PORT NETWORKS

8.3 HYBRID PARAMETERS Reconsider the two-port network of Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 at port 1 can be expressed in terms of current I1 at port 1 and voltage V2 at port 2: V1 = h11 I1 + h12 V2 (8.3.1) Similarly, we can write I2 in terms of I1 and V2 : I2 = h21 I1 + h22 V2

(8.3.2)

Since V1 and V2 are in volts while I1 and I2 are in amperes, parameter h11 must be in ohms, h12 and h21 must be dimensionless, and h22 must be in siemens. Therefore, these are called the hybrid parameters. Using the matrix representation, we can write h11 h12 I1 V1 = (8.3.3) I2 h21 h22 V2 Hybrid parameters are especially important in transistor circuit analysis. These parameters are determined as follows. If port 2 has a short circuit, V2 will be zero. In this condition, (8.3.1) and (8.3.2) give V1 h11 = (8.3.4) I1 V2 =0 and h21 =

I2 I1 V2 =0

Similarly, with a source connected at port 2 while port 1 is open, V1 h12 = V2 I1 =0 and h22

I2 = V2 I1 =0

(8.3.5)

(8.3.6)

(8.3.7)

Thus, parameters h11 and h21 represent the input impedance and the forward current gain, respectively, when a short circuit is at port 2. Similarly, h12 and h22 represent the reverse voltage gain and the output admittance, respectively, when port 1 has an open circuit. Because of this mix, these are called hybrid parameters. In transistor circuit analysis, these are generally denoted by hi , hf , hr , and h0 , respectively.

297

HYBRID PARAMETERS 12 Ω I1

3Ω

+

+

V1

6Ω

I2

V2 −

−

Figure 8.12

Two-port network for Example 8.9..

Example 8.9 Find the hybrid parameters of the two-port network shown in Figure 8.12. SOLUTION With a short circuit at port 2, 6×3 V1 = I1 12 + = 14I1 6+3 and using the current-divider rule, we find that I2 = −

6 2 I1 = − I1 6+3 3

Therefore, from (8.3.4) and (8.3.5), h11

V1 = = 14 I1 V2 =0

and h21 =

I2 2 =− I1 V2 =0 3

Similarly, with a source connected at port 2 while port 1 has an open circuit, V2 = (3 + 6)I2 = 9I2 and V1 = 6I2 because there is no current flowing through a 12- resistor. Hence, from (8.3.6) and (8.3.7), V1 6I2 2 h12 = = = V2 I1 =0 9I2 3 and h22

I2 1 = = S V2 I1 =0 9

298

Thus,

TWO-PORT NETWORKS

h11 h21

h12 h22

=

2 3 1 9

14 − 23

S

8.4 TRANSMISSION PARAMETERS Reconsider the two-port network of Figure 8.1. Since the network is linear, the superposition principle can be applied. Assuming that it contains no independent sources, voltage V1 and current I1 at port 1 can be expressed in terms of current I2 and voltage V2 at port 2 as follows: V1 = AV2 − BI2

(8.4.1)

Similarly, we can write I1 in terms of I2 and V2 as follows: I1 = CV2 − DI2

(8.4.2)

Since V1 and V2 are in volts while I1 and I2 are in amperes, parameters A and D must be dimensionless, B must be in ohms, and C must be in siemens. Using the matrix representation, (8.4.2) can be written as follows:

V1 I1

A = C

B D

V2 −I2

(8.4.3)

Transmission parameters (also known as elements of chain matrix ) are especially important for analysis of circuits connected in cascade. These parameters are determined as follows. If port 2 has a short circuit, V2 will be zero. Under this condition, (8.4.1) and (8.4.2) give V1 B= (8.4.4) −I2 V2 =0 and

I1 D= −I2 V2 =0

(8.4.5)

Similarly, with a source connected at the port 1 while port 2 is open, we find that V1 A= (8.4.6) V2 I2 =0 and

I1 C= V2 I2 =0

(8.4.7)

299

TRANSMISSION PARAMETERS

I1

1Ω +

+

V1

V2

−

−

Figure 8.13

I2

Network for Example 8.10.

Example 8.10 Determine the transmission parameters of the network shown in Figure 8.13. SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), I2 = −I1 and V1 = I1 V Therefore, from (8.4.4) and (8.4.5), B= and

V1 = 1 −I2 V2 =0

I1 D= =1 −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), V2 = V1 and I1 = 0 Now, from (8.4.6) and (8.4.7),

V1 =1 A= V2 I2 =0

and C=

I1 =0 V2 I2 =0

Hence, the transmission matrix of this network 1 A B = 0 C D

is 1 1

Example 8.11 Determine the transmission parameters of the network shown in Figure 8.14. SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), I2 = −I1 and V1 = 0V

300

TWO-PORT NETWORKS I1

+

+ jω S

V1 −

Figure 8.14

I2

V2 −

Network for Example 8.11.

Therefore, from (8.4.4) and (8.4.5), V1 = 0 B= −I2 V2 =0 and D=

I1 =1 −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), A V2 = V1 and I1 = j ωV1 Now, from (8.4.6) and (8.4.7), V1 =1 A= V2 I2 =0 and

I1 C= = jω V2 I2 =0

Hence, transmission matrix of this network is 1 A B = jω C D

S

0 1

Example 8.12 Determine the transmission parameters of the network shown in Figure 8.15. 1Ω I1

1Ω

+ V1 −

Figure 8.15

+ jω S

V2 −

Network for Example 8.12.

I2

301

TRANSMISSION PARAMETERS

SOLUTION With a source connected at port 1 while port 2 has a short circuit (so that V2 is zero), we find that 1 2 + jω V1 = 1 + I1 = I1 1 + jω 1 + jω and 1 1 jω I2 = − I1 = − I1 1/j ω + 1 1 + jω Therefore, from (8.4.4) and (8.4.5), V1 = 2 + jω B= −I2 V2 =0 and

I1 D= = 1 + jω −I2 V2 =0

Similarly, with a source connected at port 1 while port 2 is open (so that I2 is zero), 1 1 + jω I1 = I1 V1 = 1 + jω jω and V2 =

1 I1 jω

Now, from (8.4.6) and (8.4.7), V1 = 1 + jω A= V2 I2 =0 and C= Hence,

A C

B D

I1 = jω V2 I2 =0

1 + jω = jω

S

2 + jω 1 + jω

Example 8.13 Find the transmission parameters of the transmission line shown in Fig. 8.16. SOLUTION Assume that a source is connected at port 1 while the other port has a short circuit. If Vin is incident voltage at port 1, it will be Vin e−γl at port 2.

302

TWO-PORT NETWORKS l I1

+

+ Z0, γ

V1

−

Figure 8.16

I2

V2

−

Transmission line for Example 8.13.

Since the reflection coefficient of the short circuit is −1, the reflected voltage at this port is 180◦ out of phase with the incident voltage. Therefore, the reflected voltage reaching at port 1 is −Vin e−2γl . Hence, V1 = Vin − Vin e−2γl V2 = 0 Vin (1 + e−2γl ) I1 = Z0 and I2 = −

2Vin −γl e Z0

Therefore, from (8.4.4) and (8.4.5), V1 Z0 eγl − e−γl −2λl B= = (1 − e ) = Z 0 −I2 V2 =0 2e−γl 2

= Z0 sinh γl

and

I1 1 + e−2λl eγl + e−γl D= = = = cosh γl −I2 V2 =0 2e−λl 2

Now assume that port 2 has an open circuit while the source is still connected at port 1. If Vin is incident voltage at port 1, Vin e−γl is at port 2. Since the reflection coefficient of an open circuit is +1, the reflected voltage at this port is equal to the incident voltage. Therefore, the reflected voltage reaching at port 1 is Vin e−2γl . Hence, V1 = Vin + Vin e−2γl V2 = 2Vin e−γl Vin I1 = (1 − e−2γl ) Z0 and I2 = 0

303

TRANSMISSION PARAMETERS

Now, from (8.4.6) and (8.4.7), 1 + e−2γl V1 = = cosh γl A= V2 I2 =0 2e−γl and

I1 1 − e−2γl 1 C= = = sinh γl −γl V2 I2 =0 2Z0 e Z0

Hence, the transmission matrix of a finite-length transmission line is cosh γl Z0 sinh γl A B = 1 C D sinh γl cosh γl Z0 For a lossless line, γ = j β, and therefore it simplifies to cos βl j Z0 sin βl A B = 1 C D j sin βl cos βl Z0 An analysis of results obtained in Examples 8.10 to 8.13 indicates that the following condition holds for all four circuits: AD − BC = 1

(8.4.8)

This is because these circuits are reciprocal. In other words, if a given circuit is known to be reciprocal, (8.4.8) must be satisfied. Further, we find that transmission parameter A is equal to D in all four cases. This always happens when a given circuit is reciprocal. In Example 8.11, A and D are real, B is zero, and C is imaginary. For a lossless line in Example 8.13, A and D simplify to real numbers while C and D become purely imaginary. This characteristic of the transmission parameters is associated with any lossless circuit. A comparison of the circuits in Examples 8.10 to 8.12 reveals that the two-port network of Example 8.12 can be obtained by cascading that of Example 8.10 on the two sides of Example 8.11, as shown in Figure 8.17. 1Ω I1

1Ω +

+

V1 −

V2 −

jω S 8.10

Figure 8.17

8.11

8.10

Two-port network for Example 8.12.

I2

304

TWO-PORT NETWORKS

Therefore, the chain (or transmission) matrix for the network shown in Example 8.12 can be obtained after multiplying three chain matrices as follows: 1 1 1 0 1 1 1 1 1 1 · · = · 0 1 jω 1 0 1 0 1 jω jω + 1 1 + jω 2 + jω = jω 1 + jω This shows that chain matrices are convenient in the analysis and design of networks connected in cascade. 8.5 CONVERSION OF IMPEDANCE, ADMITTANCE, CHAIN, AND HYBRID PARAMETERS One type of network parameter can be converted into another via the respective defining equations. For example, the admittance parameters of a network can be found from its impedance parameters as follows. From (8.2.3) and (8.1.3), we find that −1 I1 Y11 V1 Z11 V1 Y12 Z12 = = I2 Y21 Y22 V2 Z21 Z22 V2 Hence, Y11 Y21

Y12 Y22

=

Z11 Z21

Z12 Z22

−1

=

1 Z11 Z22 − Z12 Z21

Z22 −Z21

−Z12 Z11

Similarly, (8.3.3) can be rearranged as follows: D AD − BC − I1 V1 B = B I2 V2 1 A − B B Hence, D AD − BC − Y11 Y12 B = B Y21 Y22 1 A − B B Relations between other parameters can be found following a similar procedure. These relations are given in Table 8.1.

8.6 SCATTERING PARAMETERS As illustrated in preceding sections, Z-parameters are useful in analyzing series circuits while Y -parameters simplify the analysis of parallel (shunt) connected circuits. Similarly, transmission parameters are useful for chain or cascade circuits.

305

SCATTERING PARAMETERS

TABLE 8.1 Conversions Among the Impedance, Admittance, Chain, and Hybrid Parameters Y22 Y11 Y22 − Y12 Y21 −Y12 = Y11 Y22 − Y12 Y21 −Y21 = Y11 Y22 − Y12 Y21 Y11 = Y11 Y22 − Y12 Y21

A C AD − BC = C 1 = C D = C

h11 h22 − h12 h21 h22 h12 = h22 −h21 = h22 1 = h22

Z11 =

Z11 =

Z11 =

Z12

Z12

Z12

Z21 Z22

Z22 Z11 Z22 − Z12 Z21 −Z12 = Z11 Z22 − Z12 Z21 −Z21 = Z11 Z22 − Z12 Z21 Z11 = Z11 Z22 − Z12 Z21

Z21 Z22

D B −(AD − BC) = B −1 = B A = B

Z21 Z22

1 h11 −h12 = h11 h21 = h11 h11 h22 − h12 h21 = h11

Y11 =

Y11 =

Y11 =

Y12

Y12

Y12

Y21 Y22

Z11 Z21 Z11 Z22 − Z12 Z21 B= Z21 1 C= Z21 Z22 D= Z21 A=

Z11 Z22 − Z12 Z21 Z22 Z12 = Z22 −Z21 = Z22 1 = Z22

Y21 Y22

−Y22 Y21 −1 B= Y21 −(Y11 Y22 − Y12 Y21 ) C= Y21 −Y11 D= Y21 A=

1 Y11 −Y12 = Y11 Y21 = Y11 Y11 Y22 − Y12 Y21 = Y11

h11 =

h11 =

h12

h12

h21 h22

h21 h22

Y21 Y22

−(h11 h22 − h12 h21 ) h21 −h11 B= h21 −h22 C= h21 −1 D= h21 A=

B D AD − BC = D −1 = D C = D

h11 = h12 h21 h22

306

TWO-PORT NETWORKS

However, the characterization procedure of these parameters requires an open or short circuit at the other port. This extreme reflection makes it very difficult (and in certain cases, impossible) to determine the parameters of a network at radio and microwave frequencies. Therefore, a new representation based on traveling waves is defined. This is known as the scattering matrix of the network. Elements of this matrix are known as scattering parameters. Figure 8.18 shows a network along with incident and reflected waves at its two ports. We adopt a convention of representing the incident wave by ai and the reflected wave by bi at the ith port. Hence, a1 is an incident wave, while b1 is a reflected wave at port 1. Similarly, a2 and b2 represent incident and reflected waves at port 2, respectively. Assume that a source is connected at port 1 that produces the incident wave a1 . A part of this wave is reflected back at the input (due to impedance mismatch), while the remaining signal is transmitted through the network. It may change in magnitude as well as in phase before emerging at port 2. Depending on the termination at this port, part of the signal is reflected back as input to port 2. Hence, reflected wave b1 depends on incident signals a1 and a2 at the two ports. Similarly, emerging wave b2 also depends on a1 and a2 . Mathematically, b1 = S11 a1 + S12 a2

(8.6.1)

b2 = S21 a1 + S22 a2

(8.6.2)

Using the matrix notation, we can write b1 S11 = b2 S21 or

S12 S22

a1 a2

(8.6.3)

[b] = [S][a]

(8.6.4)

where [S] is called the scattering matrix of the two-port network; Sij are known as the scattering parameters of this network, and ai represents the incident wave at the ith port and bi represents the reflected wave at the ith port. If port 2 is matched terminated while a1 is incident at port 1, a2 is zero. In this condition, (8.6.1) and (8.6.2) give b1 S11 = (8.6.5) a1 a2 =0

a1 Port 1 b 1

Two-port network

b2

Port 2

a2

Figure 8.18 Two-port network with associated incident and reflected waves.

SCATTERING PARAMETERS

and S21

b2 = a1 a2 =0

307

(8.6.6)

Similarly, with a source connected at port 2 while port 1 is terminated by a matched load, we find that b1 S12 = (8.6.7) a2 a1 =0 and S22

b2 = a2 a1 =0

(8.6.8)

Hence, Sii is the reflection coefficient i at the ith port when the other port is matched terminated. Sij is the forward transmission coefficient of the j th port if i is greater than j , whereas it represents the reverse transmission coefficient if i is less than j with the other port terminated by a matched load. We have not yet defined ai and bi in terms of voltage, current, or power. To that end, we write steady-state total voltage and current at the ith port as follows: Vi = Viin + Viref

(8.6.9)

and Ii =

1 (V in − Viref ) Z0i i

(8.6.10)

where superscripts “in” and “ref” represent the incident and reflected voltages, respectively. Z0i is the characteristic impedance at the ith port. Equations (8.6.9) and (8.6.10) can be solved to find incident and reflected voltages in terms of total voltage and current at the ith port. Hence, Viin = 12 (Vi + Z0i Ii )

(8.6.11)

Viref = 12 (Vi − Z0i Ii )

(8.6.12)

and

Assuming both of the ports to be lossless so that Z0i is a real quantity, the average power incident at the ith port is in ∗ 1 1 in in ∗ 1 in in Vi = |V in |2 (8.6.13) Pi = Re Vi (Ii ) = Re Vi 2 2 Z0i 2Z0i i and average power reflected from the ith port is ref ∗ 1 1 ref ref ∗ 1 ref ref Vi = |V ref |2 Pi = Re Vi (Ii ) = Re Vi 2 2 Z0i 2Z0i i

(8.6.14)

308

TWO-PORT NETWORKS

The ai and bi are defined in such a way that the squares of their magnitudes represent the power flowing in respective directions. Hence, Viin 1 Vi + Z0i I i 1 Vi = + Z0i Ii = √ (8.6.15) ai = √ √ √ 2 2Z 0i 2Z0i Z0i 2 2 and V ref 1 = bi = √ i 2 2Z 0i

Vi − Z0i I i √ 2Z0i

1 = √ 2 2

Vi − Z0i Ii √ Z0i

(8.6.16)

Therefore, units of ai and bi are √

√ volt watt = √ = ampere · ohm ohm

Power available from the source, Pavs , at port 1 is Pavs = |a1 |2 power reflected from port 1, Pref , is Pref = |b1 |2 and power delivered to the port (and hence to the network), Pd , is Pd = Pavs − Pref = |a1 |2 − |b1 |2 Consider the circuit arrangement shown in Figure 8.19. There is a voltage source VS1 connected at port 1 while port 2 is terminated by load impedance ZL . The source impedance is ZS . Various voltages, currents, and waves are as depicted at the two ports of this network. Further, it is assumed that the characteristic impedances at port 1 and port 2 are Z01 and Z02 , respectively. Input impedance Z1 at port 1 of the network is defined as the impedance across its terminals when port 2 is terminated by load ZL while source VS1 along with ZS are disconnected. Similarly, output impedance Z2 at port 2 of the network is defined as the impedance across its terminals with load ZL disconnected and voltage source VS1 replaced by a short circuit. Hence, source impedance ZS terminates ZS VS1

I1 a1 Port 1

b1

+ V1 −

Figure 8.19 terminated.

+ Two-port network

V2 −

I2 b2

Port 2

a2 ZL

Two-port network with a voltage source connected at port 1 and port 2 is

SCATTERING PARAMETERS

309

port 1 of the network in this case. Input impedance Z1 and output impedance Z2 are responsible for input reflection coefficient 1 and output reflection coefficient 2 , respectively. Hence, the ratio of b1 to a1 represents 1 while that of b2 to a2 is 2 . For the two-port network, we can write b1 = S11 a1 + S12 a2

(8.6.17)

b2 = S21 a1 + S22 a2

(8.6.18)

and

The load reflection coefficient L is L =

ZL − Z 02 a2 = ZL + Z02 b2

(8.6.19)

Note that b2 leaves port 2, and therefore it is incident on the load ZL . Similarly, the wave reflected back from the load enters port 2 as a2 . The source reflection coefficient S is found as S =

Z S − Z01 a1 = ZS + Z01 b1

(8.6.20)

Since b1 leaves port 1 of the network, it is the incident wave on ZS , while a1 is the reflected wave. The input and output reflection coefficients are 1 =

Z1 − Z01 b1 = Z1 + Z01 a1

(8.6.21)

2 =

Z2 − Z02 b2 = Z2 + Z02 a2

(8.6.22)

and

Dividing (8.6.17) by a1 and then using (8.6.21), we find that Z1 − Z01 a2 b1 = 1 = = S11 + S12 a1 Z1 + Z01 a1

(8.6.23)

Now, dividing (8.6.18) by a2 and then combining with (8.6.19), we get 1 a1 1 − S22 L a1 b2 = S22 + S21 = ⇒ = a2 a2 L a2 S21 L

(8.6.24)

From (8.6.23) and (8.6.24), 1 = S11 +

S12 S21 L 1 − S22 L

(8.6.25)

310

TWO-PORT NETWORKS

If a matched load is terminating port 2, then L = 0 and (7.6.25) simplifies to 1 = S11 . Similarly, from (8.6.18) and (8.6.22), Z2 − Z02 a1 b2 = 2 = = S22 + S21 a2 Z2 + Z02 a2

(8.6.26)

From (8.6.17) and (8.6.20), we have 1 a2 1 − S11 S a2 b1 = S11 + S12 = ⇒ = a1 a1 S a1 S12 S

(8.6.27)

Substituting (8.6.27) into (8.6.26), we get 2 = S22 +

S21 S12 S 1 − S11 S

(8.6.28)

If ZS is equal to Z01 , port 1 is matched and S = 0. Therefore, (8.6.28) simplifies to 2 = S22 Hence, S11 and S22 can be found by evaluating the reflection coefficients at respective ports while the other port is matched terminated. Let us determine the other two parameters, S21 and S12 , of the two-port network. Starting with (8.6.6) for S21 , we have S21

b2 = a1 a2 =0

(8.6.6)

Now, a2 is found from (8.6.15) with i as 2 and forcing it to zero, we get a2 =

1 2

V2 + Z02 I2 √ 2Z02

= 0 ⇒ V2 = −Z02 I2

(8.6.29)

Substituting (8.6.29) in the expression for b2 that is obtained from (8.6.16) with i as 2, we find that 1 b2 = 2

V2 − Z02 I2 √ 2Z02

= −I2

Z02 2

(8.6.30)

An expression for a1 is obtained from (8.6.15) with i = 1. It simplifies for ZS equal to Z01 as follows: 1 a1 = 2

V1 + Z01 I1 √ 2Z01

VS1 = √ 2 2Z01

(8.6.31)

311

SCATTERING PARAMETERS

S21 is obtained by substituting (8.6.30) and (8.6.31) into (8.6.6) as follows: √ √ b2 − Z02 I2 / 2 2V2 Z01 S21 = = = (8.6.32) √ a1 a2 =0 VS1 Z02 VS1 /2 2Z01 Following a similar procedure, S12 may be found as b1 2V1 Z02 = S12 = a2 a1 =0 VS2 Z01

(8.6.33)

An analysis of S-parameters indicates that |b1 |2 Pavs − Pd 2 = |S11 | = 2 |a1 | a2 =0 Pavs

(8.6.34)

where Pavs is power available from the source and Pd is power delivered to port 1. These two powers will be equal if the source impedance is conjugate of Z1 ; that is, the source is matched with port 1. Similarly, from (8.6.32), √ √ PAVN Z02 (I2 / 2)2 Z02 (I2 / 2)2 2 = = 1 (8.6.35) |S21 | = 1 √ √ 2 2 Pavs ( 4 Z01 )(VS1 / 2) [(1/2Z01 )(VS1 / 2) ] 2 where PAVN is power available at port 2 of the network. It will be equal to power delivered to a load that is matched to the port. This power ratio of (8.6.35) may be called the transducer power gain. Following a similar procedure, it may be found that |S22 |2 represents the ratio of power reflected from port 2 to power available from the source at port 2, while port 1 is terminated by a matched load ZS and |S12 |2 represents a reverse transducer power gain. Shifting the Reference Planes Consider the two-port network shown in Figure 8.20. Assume that ai and bi are incident and reflected waves, respectively, at unprimed reference planes of the ith A

A′

B

x1

x2 a1 Port 1

b2

Two-port network b1

Port 2

a2

l1 A′

B′

l2 A

B

Figure 8.20 Two-port network with two reference planes on each side.

B′

312

TWO-PORT NETWORKS

port. We use unprimed S-parameters for this case. Next, consider that plane A–A is shifted by a distance l1 to A –A . At this plane, a1 and b1 represent inward- and outward-traveling waves, respectively. Similarly, a2 and b2 represent inward- and outward-traveling waves, respectively, at plane B –B . We denote the scattering parameters at primed planes by a prime on each as well. Hence, b1 S11 S12 a1 = (8.6.36) b2 S21 S22 a2 and

b1 b2

=

S11 S21

S12 S22

a1 a2

(8.6.37)

Wave b1 is delayed in phase by βl1 as it travels from A to A . This means that b1 is ahead in phase with respect to b1 . Hence, b1 = b1 ej βl1

(8.6.38)

Wave a1 comes from A to plane A. Therefore, it has a phase delay of βl1 with respect to a1 . Mathematically, a1 = a1 e−j βl1

(8.6.39)

On the basis of similar considerations at port 2, we can write b2 = b2 ej βl2

(8.6.40)

a2 = a2 e−j βl2

(8.6.41)

and

Substituting for b1 , a1 , b2 , and a1 from (8.6.38) to (8.6.41) into (8.6.36), we get

b1 ej βl1 b2 ej βl2

S11 = S21

S12 S22

a1 e−j βl1

(8.6.42)

a2 e−j βl2

We can rearrange this equation as follows:

b1 b2

=

S11 e−j 2βl1

S12 e−j β(l1 +l2 )

S21 e−j β(l1 +l2 )

S22 e−j 2βl2

a1 a2

(8.6.43)

Now, on comparing (8.6.43) with (8.6.37), we find that

S11 S21

S12 S22

S11 e−j 2βl1 = S21 e−j β(l1 +l2 )

S12 e−j β(l1 +l2 ) S22 e−j 2βl2

(8.6.44)

313

SCATTERING PARAMETERS I1

I2

Two-port network

V1

Figure 8.21

V2

Network for Example 8.14.

Following a similar procedure, one can find the other relation as follows: j 2βl1 j β(l1 +l2 ) S12 e S11 e S11 S12 (8.6.45) = j β(l1 +l2 ) j 2βl2 S21 S22 S21 e S22 e Example 8.14 The total voltages and currents at two ports of a network are found as follows: V1 = 10 0◦ V, I1 = 0.1 40◦ A, V2 = 12 30◦ V, and I2 = 0.15 100◦ A. Determine the incident and reflected voltages, assuming that the characteristic impedance is 50 at both its ports (see Figure 8.21). SOLUTION From (8.6.11) and (8.6.12) with i = 1, we find that ◦

◦

◦

◦

V1in = 12 (10 0 + 50 × 0.1 40 ) = 6.915 + j 1.607 V and V1ref = 12 (10 0 − 50 × 0.1 40 ) = 3.085 − j 1.607 V Similarly, with i = 2, incident and reflected voltages at port 2 are found to be ◦

◦

◦

◦

V2in = 12 (12 30 + 50 × 0.15 100 ) = 4.545 + j 6.695 V and V2ref = 12 (12 30 − 50 × 0.15 100 ) = 5.845 − j 0.691 V Example 8.15 Find the S-parameters of a series impedance Z connected between the two ports, as shown in Figure 8.22.

Z Z0

Z0

Figure 8.22

Setup for Example 8.15.

314

TWO-PORT NETWORKS

SOLUTION From (8.6.25), with L = 0 (i.e., port 2 is terminated by a matched load), we find that S11 = 1 |a2 =0 =

(Z + Z0 ) − Z0 Z = (Z + Z0 ) + Z0 Z + 2Z0

Similarly, from (8.6.28), with S = 0 (i.e., port 1 is terminated by a matched load), (Z + Z0 ) − Z0 Z = S22 = 2 |a1 =0 = (Z + Z0 ) + Z0 Z + 2Z0 S21 and S12 are determined from (8.6.32) and (8.6.33), respectively. For evaluating S21 , we connect a voltage source VS1 at port 1 while port 2 is terminated by Z0 , as shown in Figure 8.23. The source impedance is Z0 . Using the voltage-divider formula, we can write V2 =

Z0 VS1 Z + 2Z0

Hence, from (8.6.32), S21 =

2V2 2Z0 2Z0 + Z − Z Z = = =1− VS1 Z + 2Z0 Z + 2Z0 Z + 2Z0

For evaluating S12 , we connect a voltage source VS2 at port 2 while port 1 is terminated by Z0 , as shown in Figure 8.24. The source impedance is Z0 . Using the voltage-divider rule again, we find that V1 =

Z0 VS2 Z + 2Z0

Z0

Z

VS1 ~

V2

V1

Figure 8.23

Setup to determine S21 for Example 8.15.

Z Z0

V1

Figure 8.24

Z0

Z0 V2

~ VS 2

Setup to determine S12 for Example 8.15.

315

SCATTERING PARAMETERS

Now, from (8.6.33), S12 =

2V1 2Z0 2Z0 + Z − Z Z = = =1− VS2 Z + 2Z0 Z + 2Z0 Z + 2Z0

Therefore,

S11 S21

=

1 1 − 1

1 =

Z Z + 2Z0

S12 S22

where

1 − 1 1

An analysis of the S-parameters indicates that S11 is equal to S22 . This is because the given two-port network is symmetrical. Further, its S12 is equal to S21 as well. This happens because this network is reciprocal. Consider a special case where the series impedance Z is a purely reactive element, which means that Z is equal to j X. In that case, 1 can be written as 1 =

jX j X + 2Z0

Therefore, 2 2 jX + 1 − j X + 2Z 0 0

jX |S11 | + |S21 | = j X + 2Z 2

2

=

X2 (2Z0 )2 + =1 X2 + (2Z0 )2 X2 + (2Z0 )2

Similarly, |S12 |2 + |S22 |2 = 1 On the other hand, ∗ ∗ + S21 S22 = S11 S12

jX 2Z0 2Z0 −j X × + × =0 j X + 2Z0 −j X + 2Z0 j X + 2Z0 −j X + 2Z0

and ∗ ∗ S12 S11 + S22 S21 =

2Z0 −j X jX 2Z0 × + × =0 j X + 2Z0 −j X + 2Z0 j X + 2Z0 −j X + 2Z0

These characteristics of the scattering matrix can be summarized as follows:

∗ Sij Sik

= δj k =

1 0

if j = k otherwise

316

TWO-PORT NETWORKS

Z0

Z0

Y

Figure 8.25

Network for Example 8.16.

When the elements of a matrix satisfy this condition, it is called a unitary matrix. Hence, the scattering matrix of a reactance circuit is unitary. Example 8.16 Find the S-parameters of a shunt admittance Y connected between two ports, as shown in Figure 8.25. SOLUTION From (8.6.25), S11 = 1 |L =0 = =

Z1 − Z01 1/Y1 − 1/Y0 = Z1 + Z01 1/Y1 + 1/Y0 Y0 − Y1 Y0 − (Y + Y0 ) −Y = = Y0 + Y1 Y0 + (Y + Y0 ) 2Y0 + Y

Similarly, from (8.6.28), S22 = 2 |S =0 = =

Z2 − Z02 1/Y2 − 1/Y0 = Z2 + Z02 1/Y2 + 1/Y0 Y0 − Y2 Y0 − (Y + Y0 ) −Y = = Y0 + Y2 Y0 + (Y + Y0 ) 2Y0 + Y

For evaluating S21 , voltage source VS1 is connected at port 1, and port 2 is terminated by a matched load, as shown in Figure 8.26. Using the voltage-divider rule, V2 =

1/(Y + Y0 ) 1 Y0 VS1 = VS1 = VS1 Z0 + 1/(Y + Y0 ) Z0 (Y + Y0 ) + 1 Y + 2Y0

Now, from (8.6.32), S21

2V2 2Y0 Y = = =1− = 1 + 1 VS1 L =0 Y + 2Y0 Y + 2Y0

Z0 VS1 ~

Figure 8.26

V1

Y

V2

Z0

Setup to determine S21 for Example 8.16.

317

SCATTERING PARAMETERS

Similarly, when VS2 in series with Z0 is connected at port 2 while port 1 is match-terminated, S12 can be found from (8.6.33) as follows: 2V1 2Y0 Y = =1− = 1 + 1 S12 = VS2 S =0 Y + 2Y0 Y + 2Y0 where 1 = −

Y Y + 2Y0

As expected, S11 = S22 and S12 = S21 , because this circuit is symmetrical as well as reciprocal. Further, for Y = j B, the network becomes lossless. In that case, |S11 |2 + |S21 |2 =

B2 (2Y0 )2 + =1 B 2 + (2Y0 )2 B 2 + (2Y0 )2

and ∗ ∗ + S21 S22 = S11 S12

−jB 2Y0 2Y0 jB × + × =0 jB + 2Y0 −jB + 2Y0 jB + 2Y0 −jB + 2Y0

Hence, with Y as jB, the scattering matrix is unitary. In fact, it can easily be proved that the scattering matrix of any lossless two-port network is unitary. Example 8.17 An ideal transformer is designed to operate at 500 MHz. It has 1000 turns on its primary side and 100 turns on its secondary side. Assuming that it has a 50- connector on each side, determine its S-parameters (see Figure 8.27). SOLUTION For an ideal transformer, V1 −I2 = =n V2 I1

and

V1 (−I2 ) V1 /I1 Z1 = = n2 = V2 I 1 V2 /(−I2 ) Z2

Now, following the procedure used in preceding examples, we find that b1 Z0 n2 − Z0 n2 − 1 99 S11 = = | = = = 1 a2 =0 2 2 a1 a2 =0 Z0 n + Z0 n +1 101

I1 Port 1

n :1 + + V2 −

V1 − 1000 : 100

Figure 8.27 Transformer for Example 8.17.

I2 Port 2

318

TWO-PORT NETWORKS

and voltage V1 on the primary side of the transformer due to a voltage source VS1 with its internal impedance Z0 can be determined by the voltage-divider rule as follows: V1 =

Z0 n2 VS1 Z0 n2 + Z0

Hence, S21

V1 n2 nV2 = 2 = VS1 n +1 VS1

⇒

2V2 2n 20 = = 2 = VS1 a2 =0 n +1 101

Similarly, S22 and S12 are determined after connecting a voltage source VS2 with its internal impedance Z0 at port 2. Port 1 is terminated by a matched load this time. Therefore, b2 (Z0 /n2 ) − Z0 1 − n2 99 = 2 |a1 =0 = = =− S22 = 2 2 a2 a1 =0 (Z0 /n ) + Z0 1+n 101 and V2 =

Z0 /n2 V2 1/n2 V1 /n 1 VS2 ⇒ = = = 2 2 2 (Z0 /n ) + Z0 VS2 (1/n ) + 1 1+n VS2

Hence, S12 Thus,

2V1 2n 20 = = 2 = VS2 a1 =0 n +1 101 S11 S21

S12 S22

=

99 101 20 101

20 101 99 − 101

This time, S11 is different from S22 because the network is not symmetrical. However, S12 is equal to S21 because it is reciprocal. Further, an ideal transformer is lossless. Hence, its scattering matrix must be unitary. Let us verify if that is the case. 10,201 99 2 20 2 2 2 + = |S11 | + |S21 | = =1 101 101 10,201 20 2 99 2 10,201 + − = =1 |S12 |2 + |S22 |2 = 101 101 10,201 20 20 99 99 ∗ ∗ × + × − S11 S12 + S21 S22 = =0 101 101 101 101 and ∗ S12 S11

+

∗ S22 S21

99 99 20 20 × + − =0 = × 101 101 101 101

Hence, the scattering matrix is indeed unitary.

319

SCATTERING PARAMETERS l I1 a1

I2 +

+

b2

Z0, γ V1

V2

−

−

b1

a2

Figure 8.28 Transmission line network for Example 8.18.

Example 8.18 Find the scattering parameters of the transmission line network shown in Figure 8.28. SOLUTION With port 2 match-terminated, wave a1 entering port 1 emerges as a1 e−γl at port 2. This wave is absorbed by the termination, making a2 zero. Further, b1 is zero as well because Z1 is equal to Z0 in this case. Hence, b2 = a1 e−γl Similarly, when a source is connected at port 2 while port 1 is match-terminated, we find that a1 and b2 are zero and b1 = a2 e−γl Hence,

S11 S21

S12 S22

=

0 e−γl

e−γl 0

As expected, S11 = S22 and S12 = S21 , because this circuit is symmetrical as well as reciprocal. Further, for γ = j β, the network becomes lossless. In that case, 0 e−j βl S11 S12 = −j βl S21 S22 e 0 Obviously, it is a unitary matrix now. Example 8.19 Find the scattering parameters of the two-port network shown in Figure 8.29. SOLUTION With port 2 matched-terminated by a 50- load, impedance Z1 at port 1 is Z1 = j 50 +

50(−j 25) = j 50 + 10 − j 20 = 10 + j 30 50 − j 25

320

TWO-PORT NETWORKS

j50 Ω

Port 1

Port 2

Z01 = 50 Ω

Z02 = 50 Ω

−j 25 Ω

Figure 8.29

Two-port network for Example 8.19.

Therefore, S11 = 1 |a2 =0 =

Z1 − Z01 10 + j 30 − 50 ◦ = = 0.74536 116.565 Z1 + Z01 10 + j 30 + 50

Similarly, output impedance Z2 at port 2 when port 1 is match-terminated by a 50- load is (50 + j 50)(−j 25) Z2 = = 10 − j 30 50 + j 50 − j 25 and from (8.6.32), S22 = 2 |a1 =0 =

Z2 − Z02 10 − j 30 − 50 = = 0.74536 Z1 + Z02 10 − j 30 + 50

◦

− 116.565

Now, let us connect a voltage source VS1 with its impedance at 50 at port 1 while port 2 is matched, as shown in Figure 8.30. Using the voltage-divider rule, we find that 10 − j 20 V2 = VS1 50 + j 50 + 10 − j 20 Therefore, S21

2V2 10 − j 20 = =2 = 0.66666 VS1 a2 =0 60 + j 30

50 Ω

◦

− 90

V2 j50 Ω

VS1 ~

Z01 = 50 Ω

Port 2 −j25 Ω

Z02 = 50 Ω

Port 1

Figure 8.30

Setup to determine S21 for Example 8.19.

50 Ω

321

SCATTERING PARAMETERS

V1

50 Ω

V2 j50 Ω

Z01 = 50 Ω

Z02 = 50

−j 25 Ω

~

Port 2

Port 1

VS2

Figure 8.31 Setup to determine S12 for Example 8.19.

Now connect the voltage source at port 2 while port 1 is match-terminated as shown in Figure 8.31. Using the voltage-divider rule again, we find that V2 =

10 − j 30 VS2 = 0.4714 50 + 10 − j 30

◦

− 45 V

Using it one more time, V1 is found as follows: V1 =

50 1 V2 = 0.4714 50 + j 50 1 + j1

◦

− 45 = 0.33333

◦

− 90 V

and now from (8.6.33), we get S12

2V1 = = 2 × 0.33333 VS2 a1 =0

◦

− 90 = 0.66666

◦

− 90

Hence,

S11 S21

S12 S22

0.74536 116.565◦ = 0.66666 − 90◦

0.66666 0.74536

− 90◦ − 116.565◦

In this case, S11 is different from S22 because the network is not symmetrical. However, S12 is equal to S21 because the network is reciprocal. Further, this network is lossless. Hence, its scattering matrix must be unitary. It can be verified as follows: |S11 |2 + |S21 |2 = (0.74536)2 + (0.66666)2 = 0.99999 = 1 |S12 |2 + |S22 |2 = (0.66666)2 + (0.74536)2 = 0.99999 = 1 ◦

◦

∗ ∗ S11 S12 + S21 S22 = (0.74536 116.565 ) × (0.66666 90 )

+ (0.66666

◦

◦

− 90 ) × (0.74536 116.565 ) = 0

322

TWO-PORT NETWORKS

and ∗ ∗ S12 S11 + S22 S21 = (0.66666

+ (0.74536

◦

− 90 ) × (0.74536

◦

− 116.565 )

◦

◦

− 116.565 ) × (0.66666 90 )

=0 Hence, this scattering matrix is indeed unitary. Table 8.2 summarizes the characteristics of parameter matrices of reciprocal and symmetrical two-port networks. The properties of scattering parameters of lossless junctions are listed in Table 8.3. TABLE 8.2 Properties of Parameters of Reciprocal and Symmetrical Two-Port Networks Parameter Matrix Z11 Z12 Z21 Z22 Y11 Y12 Y21 Y22 A B C D S11 S12 S21 S22

TABLE 8.3

Properties Z12 = Z21 Z11 = Z22 Y12 = Y21 Y11 = Y22 AD − BC = 1 A=D S12 = S21 S11 = S22

Properties of Scattering Matrix of Lossless Junctions

Properties

Explanation

Matrix [S] is symmetrical. [S]t = [S], where [S]t is the transpose matrix of [S]. Consequently, Sij = Sj i Matrix [S] is unitary.

[S]a = [S ∗ ]t = [S]−1 , where [S]a is the adjoint matrix of [S], [S ∗ ]t is the conjugate of [S]t , and [S]−1 is the inverse matrix of [S]. Consequently, N 1 for j = k ∗ Sij Sik = δj k = 0 otherwise i=1 Therefore, N i=1

Sij Sij∗ =

N i=1

|Sij |2 = 1,

j = 1, 2, 3, . . . , N

323

CONVERSION TO AND FROM SCATTERING PARAMETERS

8.7 CONVERSION FROM IMPEDANCE, ADMITTANCE, CHAIN, AND HYBRID PARAMETERS TO SCATTERING PARAMETERS, OR VICE VERSA From (8.1.3) we can write 2 2 Vi 1 Zik =√ Zik Ik = √ Z0i 2Z0i 2Z0i k=1 k=1

=

2

Z ik

k=1

Z0i Ik 2

Z0i Ik 2

where i = 1, 2

Therefore, from (8.6.15) and (8.6.16), we have

and

2 1 Z0i ai = (Z ik + δik ) Ik 2 k=1 2

(8.7.1)

2 1 Z0i Ik (Z ik − δik ) bi = 2 k=1 2

(8.7.2)

where δik =

1 for i = k 0 otherwise

Equations (8.7.1) and (8.7.2) can be written in matrix form as follows: [a] = 12 {[Z] + [U ]}[I ]

(8.7.3)

[b] = 12 {[Z] − [U ]}[I ]

(8.7.4)

where Ik =

Z0i Ik 2

and [U ] is the unit matrix. From (8.7.3) and (8.7.4), we have

Hence,

[b] = {[Z] − [U ]}{[Z] + [U ]}−1 [a]

(8.7.5)

[S] = {[Z] − [U ]}{[Z] + [U ]}−1

(8.7.6)

324

TWO-PORT NETWORKS

TABLE 8.4 Conversion from (to) Impedance, Admittance, Chain, or Hybrid to (from) Scattering Parameters ∗ )(Z22 + Z02 ) − Z12 Z21 (Z11 − Z01 (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21 √ 2Z12 R01 R02 = (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21 √ 2Z21 R01 R02 = (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21

∗ + S11 Z01 )(1 − S22 ) + S12 S21 Z01 (Z01 (1 − S11 )(1 − S22 ) − S12 S21 √ 2S12 R01 R02 = (1 − S11 )(1 − S22 ) − S12 S21 √ 2S21 R01 R02 = (1 − S11 )(1 − S22 ) − S12 S21

S11 =

Z11 =

S12

Z12

S21

S22 =

∗ Z02 )

(Z11 + Z01 )(Z22 − − Z12 Z21 (Z11 + Z01 )(Z22 + Z02 ) − Z12 Z21

∗ ∗ (1 − Y11 Z01 )(1 + Y22 Z02 ) + Y12 Y21 Z01 Z02 (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02 √ −2Y12 R01 R02 = (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02 √ −2Y21 R01 R02 = (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02

Z21

Z22 =

∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 ∗ + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ −2S12 R01 R02 = ∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ −2S21 R01 R02 = ∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02

S11 =

Y11 =

S12

Y12

S21

S22 =

∗ Y22 Z02 )

∗ Y12 Y21 Z01 Z02

(1 + Y11 Z01 )(1 − + (1 + Y11 Z01 )(1 + Y22 Z02 ) − Y12 Y21 Z01 Z02

Y21

∗ (Z02 + S22 Z02 )(1 − S11 ) + S12 S21 Z02 (1 − S11 )(1 − S22 ) − S12 S21

Y22 =

∗ (Z01

∗ (Z01

∗ (Z01 + S11 Z01 )(1 − S22 ) + S12 S21 Z01 ∗ + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02

S11 =

A=

∗ (Z01 + S11 Z01 )(1 − S22 ) + S12 S21 Z01 √ 2S21 R01 R02

S12

B=

∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) − S12 S21 Z01 Z02 √ 2S21 R01 R02

C=

(1 − S11 )(1 − S22 ) − S12 S21 √ 2S21 R01 R02

D=

∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ 2S21 R01 R02

S21

∗ ∗ AZ02 + B − CZ01 Z02 − DZ01 AZ02 + B + CZ01 Z02 + DZ01 √ 2(AD − BC)( R01 R02 ) = AZ02 + B + CZ01 Z02 + DZ01 √ 2 R01 R02 = AZ02 + B + CZ01 Z02 + DZ01

S22 =

∗ ∗ −AZ02 + B − CZ01 Z02 + DZ01 AZ02 + B + CZ01 Z02 + DZ01

∗ (h11 − Z01 )(1 + h22 Z02 ) − h12 h21 Z02 (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02 √ 2h12 R01 R02 = (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02 √ −2h21 R01 R02 = (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02

∗ ∗ (Z01 + S11 Z01 )(Z02 + S22 Z02 ) + S12 S21 Z01 Z02 ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ 2S12 R01 R02 = ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02 √ −2S21 R01 R02 = ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02

S11 =

h11 =

S12

h12

S21

S22 =

or

∗ ∗ (Z01 + h11 )(1 − h22 Z02 ) + h12 h21 Z02 (Z01 + h11 )(1 + h22 Z02 ) − h12 h21 Z02

h21

h22 =

(1 − S11 )(1 − S22 ) − S12 S21 ∗ (1 − S11 )(Z02 + S22 Z02 ) + S12 S21 Z02

[Z] = {[U ] + [S]}{[U ] − [S]}−1

(8.7.7)

Similarly, other conversion relations can be developed. Table 8.4 lists these equations when the two ports have different complex characteristic impedances. Characteristic impedance at port 1 is assumed Z01 , with its real part as R01 . Similarly, characteristic impedance at port 2 is Z02 , with its real part as R02 .

325

SUGGESTED READING

TABLE 8.5 Conversion Between Scattering and Chain Scattering Parameters S11 =

T21 T11

T11 =

S12 =

T11 T22 − T12 T21 T11

T12

S21 =

1 T11

T21 =

S11 S21

S22 =

−T12 T11

T22 =

−(S11 S22 − S12 S21 ) S21

1 S21 −S22 = S21

8.8 CHAIN SCATTERING PARAMETERS Chain scattering parameters, also known as scattering transfer parameters or T parameters, are useful for networks in cascade. These are defined on the basis of waves a1 and b1 at port 1 as dependent variables and waves a2 and b2 at port 2 as independent variables. Hence,

a1 b1

=

T11 T21

T12 T22

b2 a2

Conversion relations of chain scattering parameters with others can easily be developed. Conversions between S- and T-parameters are listed in Table 8.5.

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Gonzalez, G., Microwave Transistor Amplifiers. Upper Saddle River, NJ: Prentice Hall, 1997. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Ramo, S., J. R. Whinnery, and T. Van Duzer, Fields and Waves in Communication Electronics. New York: Wiley, 1994. Rizzi, P. A., Microwave Engineering. Englewood Cliffs, NJ: Prentice Hall, 1988.

326

TWO-PORT NETWORKS

PROBLEMS 8.1. Determine the Z and Y matrixes for the two-port network shown in Figure P8.1. The signal frequency is 500 MHz. I1

I2 4 pF

V1

V2

Figure P8.1

8.2. Determine the Z-parameters of the two-port network shown in Figure P8.2. 20 Ω 10 Ω

10 Ω

1Ω

Figure P8.2

8.3. Determine the Z-parameters of the two-port network shown in Figure P8.3 operating at 800 MHz. 50 Ω

Port 1

50 Ω

10 pF

Figure P8.3

Port 2

327

PROBLEMS

8.4. Determine the Z-matrix for the two-port network shown in Figure P8.4. j2 Ω

j3 Ω j4 Ω

Figure P8.4

8.5. The two-port network shown in Figure P8.5 represents a high-frequency equivalent model of the transistor. If the transistor is operating at 108 rad/s, find its Z-parameters. 1 pF

V1

5 pF 100 kΩ

10 kΩ

0.01V1

V2

Figure P8.5

8.6. Determine the Y-parameters of the two-port network shown in Figure P8.6. 50 Ω

Z0 = 50 Ω

Z0 = 50 Ω

50 Ω

Figure P8.6

8.7. Find the Y-parameters of the two-port network shown in Figure P8.7. 3Ω

Port 1

5Ω

10 Ω

Figure P8.7

2Ω

20 Ω

Port 2

328

TWO-PORT NETWORKS

8.8. Measurements are performed with a two-port network as shown in Figure P8.8. The voltages and currents observed are tabulated in Table P8.8. However, there are a few blank entries in this table. Fill in those blanks and find the Y and Z parameters of this two-port network. I1

I2

~ V1

~

V2

Figure P8.8

TABLE P8.8 Experiment

V1 (V)

V2 (V)

1 2 3 4 5

20 50

0 100

100

50

I1 (A)

I2 (A)

4 −20 5

−8 −5 0

5

15

8.9. The Y-matrix of a two-port network is given as

10 −5 [Y ] = 50 20

mS

It is connected in a circuit as shown in Figure P8.9. Find the voltages V1 and V2 . 25 Ω 100 V ∠0°

V1

[Y ] = 10 −5 50 20

mS

V2

100 Ω

Figure P8.9

8.10. Determine the transmission and hybrid parameters for the networks shown in Figure P8.10.

329

PROBLEMS 20 Ω

200 Ω

200 Ω

(a)

10 Ω

10 Ω 200 + j 8 Ω

(b)

Figure P8.10

8.11. Determine the transmission parameters (ABCD) of the circuit shown in Figure P8.11 operating at ω = 106 rad/s. 1 µH

1 µF I2

I1

V1

1Ω

V2

Figure P8.11

8.12. Find the transmission matrix of the two-port network shown in Figure P8.12.

j50 Ω 50 Ω

50 Ω

Figure P8.12

330

TWO-PORT NETWORKS

8.13. Find the transmission matrix (ABCD parameters) of the circuit shown in Figure P8.13 at 4 GHz. 0.159 pF

100 Ω

220 Ω

220 Ω

Figure P8.13

8.14. Find the transmission matrix of the two-port network shown in Figure P8.14. −j50 Ω

50 Ω

j50 Ω

Port 1

Port 2

Figure P8.14

8.15. Determine the S-parameters for the network shown in Figure P8.15.

+j50 Ω

−j50 Ω Z0 = 50 Ω

50 Ω

Z0 = 50 Ω

Figure P8.15

8.16. Calculate the Z and S parameters for the circuit shown in Figure P8.16 at 4 GHz. Both ports of the network have characteristic impedance of 50 . 0.5 pF

292 Ω

17.6 Ω

Figure P8.16

292 Ω

331

PROBLEMS

8.17. Determine the S-parameters of the two-port T-network shown in Figure P8.17. 10 Ω

Z0 = 50 Ω

20 Ω

Z0 = 50 Ω

30 Ω

Figure P8.17

8.18. Determine the S-parameters of the two-port network shown in Figure P8.18.

1Ω

1Ω Z0 = 50 Ω

Z0 = 50 Ω 10 Ω

Figure P8.18

8.19. Determine the S-parameters of the two-port network shown in Figure P8.19. 50 Ω

Z0 = 50 Ω

50 Ω

Z0 = 50 Ω

Figure P8.19

8.20. Determine the S-parameters of the two-port π-network shown in Figure P8.20. 200 Ω

Z0 = 50 Ω

200 Ω

Figure P8.20

200 Ω

Z0 = 50 Ω

332

TWO-PORT NETWORKS

8.21. Scattering variables measured at a port are found as follows: a = 8 − j 5 and b = 2 + j 5. Normalizing impedance is Z0 = 50 . Calculate the corresponding voltage and current. 8.22. Scattering variables measured at a port are a = 0.85 45◦ and b = 0.25 65◦ . The normalizing impedance is Z0 = 50 . Calculate the corresponding voltage and current. 8.23. Calculate the power flow into a port in each case for the following sets of scattering variables at the port: (a) a = 3 + j 5, b = 0.2 − j 0.1 (b) a = 40 − j 30, b = 5 − j 5 8.24. Calculate the power flow into a port in each case for the following sets of scattering variables at the port: (a) a = 5 60◦ , b = 2 −45◦ (b) a = 50 0◦ , b = 10 0◦ 8.25. Calculate the scattering variables a and b in each of the following cases: (a) V = 4 − j 3 V, I = 0.3 + j 0.4 A (b) V = 10 −30◦ V, I = 0.7 70◦ A Assume that characteristic impedance at the port is 100 . 8.26. Find the reflection coefficient for each of the following sets of scattering variables: (a) a = 0.3 + j 0.4, b = 0.1 − 0.2 (b) a = −0.5 + j 0.2, b = 0 − j 0.1 (c) a = 0.5 −70◦ , b = 0.3 20◦ (d) a = 5 0◦ , b = 0.3 90◦ 8.27. The scattering parameters of a two-port network are given as follows: S11 = 0.687 107◦ , S21 = 1.72 −59◦ , S12 = 0.114 −81◦ , and S22 = 0.381 153◦ . A source of 3 mV with an internal resistance of 50 is connected at port 1. Assuming the characteristic impedance at its ports as 50 , determine the scattering variables a1 , b1 , a2 , and b2 for the following load conditions: (a) 50 and (b) 100 .

9 FILTER DESIGN

A circuit designer frequently requires filters to extract the desired frequency spectrum from a wide variety of electrical signals. If a circuit passes all signals from dc through a frequency ωc but stops the rest of the spectrum, it is known as a low-pass filter. The frequency ωc is called its cutoff frequency. Conversely, a high-pass filter stops all signals up to ωc and passes those at higher frequencies. If a circuit passes only a finite frequency band that does not include zero (dc) and infinite frequency, it is called a bandpass filter. Similarly, a bandstop filter passes all signals except a finite band. Thus, bandpass and bandstop filters are specified by two cutoff frequencies to set the frequency band. If a filter is designed to block a single frequency, it is called a notch filter. The ratio of the power delivered by a source to a load with and without a two-port network inserted in between is known as the insertion loss of that twoport. It is generally expressed in decibels. The fraction of the input power that is lost due to reflection at its input port is called the return loss. The ratio of the power delivered to a matched load to that supplied to it by a matched source is called the attenuation of that two-port network. Filters have been designed using active devices such as transistors and operational amplifiers, as well as with only passive devices (inductors and capacitors only). Therefore, these circuits may be classified as active or passive filters. Unlike passive filters, active filters can amplify the signal besides blocking the undesired frequencies. However, passive filters are economical and easy to design. Further, passive filters perform fairly well at higher frequencies. In this chapter we present the design procedure of these passive circuits. Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

333

334

FILTER DESIGN

There are two methods available to synthesize passive filters. One of them is known as the image parameter method and the other as the insertion-loss method. The former provides a design that can pass or stop a certain frequency band, but its frequency response cannot be shaped. The insertion-loss method is more powerful in the sense that it provides a specified response of the filter. Both of these techniques are described in this chapter. The chapter concludes with a design overview of microwave filters.

9.1 IMAGE PARAMETER METHOD Consider the two-port network shown in Figure 9.1. V1 and V2 represent voltages at its ports. Currents I1 and I2 are assumed as indicated in the figure. Note that I1 is entering port 1 while I2 is leaving port 2. Further, Zin is the input impedance at port 1 when Zi2 terminates at port 2. Similarly, Z0 is the output impedance with Zi1 connected at port 1. Zi1 and Zi2 are known as the image impedance of the network. Following the transmission parameter description of the two-port, we can write V1 = AV2 + BI2

(9.1.1)

I1 = CV2 + DI2

(9.1.2)

and

Therefore, the impedance Zin at its input can be found as Zin =

V1 AV2 + BI2 AZi2 + B = = I1 CV2 + DI2 CZi2 + D

(9.1.3)

Alternatively, (9.1.1) and (9.1.2) can be rearranged as follows after noting that AD —BC must be unity for a reciprocal network. Hence, V2 = DV1 − BI1

(9.1.4)

I2 = −CV1 + AI1

(9.1.5)

and

I2

Zi1

V1

Zin

A

B

C

D

Z0

Figure 9.1 Two-port network with terminations.

V2

Zi2

IMAGE PARAMETER METHOD

335

Therefore, the output impedance Z0 is found to be Z0 = −

V2 DV1 − BI1 DZi1 + B =− = I2 −CV1 + AI1 CZi1 + A

Note that Zi1 = −

(9.1.6)

V1 I1

For Zin = Zi1 and Z0 = Zi2 , equations (9.1.3) and (9.1.6) give Zi1 (CZi2 + D) = AZi2 + B

(9.1.7)

Zi2 (CZi1 + A) = DZi1 + B

(9.1.8)

and

Subtracting equation (9.1.8) from (9.1.7), we find that Zi2 =

D Zi1 A

(9.1.9)

Now, substituting (9.1.9) into (9.1.7), we have Zi1 =

AB CD

(9.1.10)

Similarly, substituting (9.1.10) into (9.1.9) yields Zi2 =

BD AC

(9.1.11)

The transfer characteristics of the network can be formulated as follows. From (9.1.1), V1 I2 B =A+B =A+ V2 V2 Zi2 or V1 = V2

√ A √ ( AD + BC) D

For a reciprocal two-port network, AD − BC is unity. Therefore, V2 = V1

D AD − BC = √ √ A AD + BC

√ D √ ( AD − BC) A

(9.1.12)

336

FILTER DESIGN

Similarly, from (9.1.2), I2 = I1

√ A √ ( AD − BC) D

(9.1.13)

Note that equation (9.1.12) is similar to (9.1.13) except that the multiplying coefficient in one is the reciprocal of the other. This coefficient may be interpreted as the transformer turns ratio. It is unity for symmetrical T and π networks. The propagation factor γ (equal to α + j β, as usual) of the network can be defined as e−γ =

√

AD −

√

BC

Since AD − BC = 1, we find that cosh γ =

√

AD

(9.1.14)

The characteristic parameters of π and T networks are summarized in Table 9.1; the corresponding low-pass and high-pass circuits are given in Table 9.2.

TABLE 9.1

Parameters of T and π Networks π-Network

T-Network Z1/2

Z1

A =1+

Z1 2Z2

B = Z1 C =

Z2

2Z2

2Z2

ABCD parameters

Z1/2

Z1 2Z2

B = Z1 +

1 Z1 + Z2 4Z22

D =1+

A =1+

Z1 2Z2

C =

1 Z2

D =1+

Z1 Z2 Z1 Z2 = 1 + Z1 /4Z2 ZiT

Image impedance

Ziπ =

Propagation constant, γ

cosh γ = 1 +

Z1 2Z2

ZiT =

Z12 4Z2

Z1 2Z2

Z1 Z1 Z2 1 + 4Z2

cosh γ = 1 +

Z1 2Z2

337

IMAGE PARAMETER METHOD

TABLE 9.2

Constant-k Filter Sections

Filter Type Low-pass

π-Section

T-Section L /2

L

L /2

C/2

C

High-pass

2C

C/2

C

2C

L

2L

2L

For a low-pass T-section as illustrated in Table 9.2, Z1 = j ωL

(9.1.15)

1 j ωC

(9.1.16)

and Z2 =

Therefore, its image impedance can be found from Table 9.1 as follows: ZiT =

L ω2 LC 1− C 4

(9.1.17)

In the case of a dc signal, the second term inside the parentheses will be zero and the resulting image impedance is generally known as the nominal impedance, Z0 . Hence, L Z0 = C Note that the image impedance goes to zero if ω2 LC/4 = 1. The frequency that satisfies this condition is known as the cutoff frequency, ωc . Hence, ωc = √

2 LC

(9.1.18)

338

FILTER DESIGN

Similarly, for a high-pass T-section, 1 j ωC

(9.1.19)

Z2 = j ωL

(9.1.20)

Z1 = and

Therefore, its image impedance is found to be ZiT =

L 1 1− C 4ω2 LC

(9.1.21)

The cutoff frequency of this circuit will be given as 1 ωc = √ 2 LC

(9.1.22)

Example 9.1 Design a low-pass constant-k T-section that has a nominal impedance of 75 and a cutoff frequency of 2 MHz. Plot its frequency response in the frequency band 100 kHz to 10 MHz. SOLUTION Since the nominal impedance Z0 must be 75 , Z0 = 75 =

L C

The cutoff frequency ωc of a low-pass T-section is given by (9.1.18). Therefore, ωc = 2π × 2 × 106 = √

2 LC

These two equations can be solved for the inductance L and capacitance C, as follows: L = 11.9366 µH and C = 2.122 nF This circuit is illustrated in Figure 9.2. Note that inductance L calculated here is twice the value needed for a T-section. Propagation constant γ of this circuit is

339

IMAGE PARAMETER METHOD

5.9683 µH

5.9683 µH

2.122 nF

Figure 9.2

Low-pass constant-k T-section.

0.1

0.2

0.5

1

2

5

10

0.1

0.2

0.5 1 2 Frequency (MHz)

5

10

0.1

0.2

0.5

2

5

10

0.2

0.5 1 2 Frequency (MHz)

5

10

0

Magnitude (dB)

−10

−20

−30

−40

1

−25

Phase (deg)

−50 −75 −100 −125 −150 −175 0.1

Figure 9.3

Frequency response of the T-section shown in Figure 9.2.

determined from the formula listed in Table 8.1. The transfer characteristics are then found as e−γ . The frequency response of the designed circuit is shown in Figure 9.3. The magnitude of the transfer function (ratio of the output to input voltages) remains constant at 0 dB for frequencies lower than 2 MHz. Therefore, the

340

FILTER DESIGN

output magnitude will be equal to the input in this range. It falls by 3 dB if the signal frequency approaches 2 MHz. It falls continuously as the frequency increases further. Note that the phase angle of the transfer function remains constant at −180◦ beyond 2 MHz. However, it shows a nonlinear characteristic in the passband. The image impedance ZiT of this T-section can be found from (9.1.17). Its characteristics (magnitude and phase angle) with frequency are displayed in Figure 9.4. The magnitude of ZiT reduces continuously as frequency is increased and becomes zero at the cutoff. The phase angle of ZiT is zero in passband, and it changes to 90◦ in the stopband. Thus, image impedance is a variable resistance in the passband, whereas it switches to an inductive reactance in the stopband. The frequency characteristics illustrated in Figures 9.3 and 9.4 are representative of any constant-k filter. There are two major drawbacks to this type of filter: 1. The signal attenuation rate after the cutoff point is not very sharp, 2. The image impedance is not constant with frequency. From a design point of view, it is important that it stay constant, at least in its passband.

350

Magnitude (Ω)

300 250 200 150 100 50 2

4 6 Frequency (MHz)

8

10

80 60 40 20

2

Figure 9.4 frequency.

4 6 Frequency (MHz)

8

10

Image impedance of the constant-k filter of Figure 9.2 as a function of

341

IMAGE PARAMETER METHOD

These problems can be remedied using the techniques described in the following section.

m-Derived Filter Sections Consider the two T-sections shown in Figure 9.5. The first network represents the constant-k filter considered in the preceding section, and the second is a new m-derived section. It is assumed that the two networks have the same image impedance. From Table 9.1 we can write ZiT

= ZiT =

Z1 Z2

Z 1 + 1 4Z2

=

Z1 Z1 Z2 1 + 4Z2

It can be solved for Z2 as follows: Z2

1 = Z1

Now assume that

Z 2 − Z12 Z1 Z2 + 1 4

(9.1.23)

Z1 = mZ1

(9.1.24)

Substituting (9.1.24) in to (9.1.23), we get Z2 =

1 − m2 Z2 + Z1 m 4m

(9.1.25)

Thus, an m-derived section is designed from the values of components determined for the corresponding constant-k filter. The value of m is selected to sharpen the attenuation at cutoff or to control the image impedance characteristics in the passband.

Z1/2

Figure 9.5

Z1/2

Z′1/2

Z′1/2

Z2

Z′2

(a)

(b)

Constant-k T-section (a) and m-derived section (b).

342

FILTER DESIGN

For a low-pass filter, the m-derived section can be designed from the corresponding constant-k filter using (9.1.24) and (9.1.25) as follows: Z1 = j ωmL

(9.1.26)

and Z2 =

1 1 − m2 j ωL + 4m j ωmC

(9.1.27)

Now we need to find its propagation constant γ and devise some way to control its attenuation around the cutoff. Expression for the propagation constant of a Tsection is listed in Table 9.1. To find γ of this T-section, we first divide (9.1.26) by (9.1.27): Z1 ω2 m2 LC 4ω2 m2 /ω2c = − = − Z2 1 − [(1 − m2 )/4]ω2 LC 1 − (1 − m2 )ω2 /ω2c where ωc = √

2 LC

(9.1.28)

(9.1.29)

Using the formula listed in Table 9.1 and (9.1.28), the propagation constant γ is found as follows: cosh γ = 1 +

2 (mω/ωc )2 Z1 =1− 2Z2 1 − (1 − m2 ) (ω/ωc )2

or cosh γ =

ω2c − ω2 − (mω)2 ω2c − (1 − m2 )ω2

(9.1.30)

Hence, the right-hand side of (9.1.30) is dependent on the frequency ω. It will go to infinity (and therefore, γ) if the following condition is satisfied: ω= √

ωc 1 − m2

= ω∞

(9.1.31)

This condition can be used to sharpen the attenuation at cutoff. A small m will place ω∞ close to ω. In other words, ω∞ is selected a little higher than ωc and the fractional value of m is then determined from (9.1.31). Z1 and Z2 of the mderived section are subsequently found using (9.1.24) and (9.1.25), respectively. Note that the image impedance of an m-derived T-section is the same as that of the corresponding constant-k network. However, it will be a function of m in the case of a π-network. Therefore, this characteristic can be utilized to design input and output networks of the filter so that the image impedance of the composite circuit stays almost constant in its passband. Further, an infinite cascade of T-sections can be considered as a π-network after splitting its shunt

343

IMAGE PARAMETER METHOD

Z1′/2

Z1′/2

Z1′/2

Z2′

Z1′/2

Z2′

(a) Z1′

2Z2′

2Z2′

(b)

Figure 9.6 Deembedding of a π-network (b) from a cascaded T-network (a).

arm as illustrated in Figure 9.6. Note that the impedance Z2 of the T-network is replaced by 2Z2 , while two halves of the series arms of the T-network give Z1 of the π-network. Image impedance Ziπ is found from Table 9.1 as follows: Ziπ =

Z1 Z2 Z1 Z2 + [(1 − m2 )/4]Z12 = ZiT ZiT

(9.1.32)

For the low-pass constant-k filter, we find from (9.1.15) to (9.1.18) that Z1 Z2 =

L = Z02 C

Z12 = −ω2 L2 = and,

ZiT = Z0 1 −

2Z0 ω ωc

ω ωc

2

2

Therefore, Ziπ =

1 − (1 − m2 ) (ω/ωc )2 Z0 1 − (ω/ωc )2

344

FILTER DESIGN 2.5

Normalized impedance

2.0

m = 0.8

1.5

m = 0.6

1.0 m = 0.4 0.5 0.0

0.2

0.4

0.6

0.8

1.0

Normalized frequency

Figure 9.7 Normalized image impedance of π-network versus normalized frequency for three values of m.

or Z iπ =

1 − (1 − m2 )ω2 (1 − ω2 )

(9.1.33)

where Z iπ = Ziπ /Z0 may be called the normalized image impedance of a πnetwork and ω = ω/ωc the normalized frequency. Equation (9.1.33) is displayed graphically in Figure 9.7. The normalized image impedance is close to unity for nearly 90% of the passband if m is kept around 0.6. Therefore, it can be used as pre- and poststages of the composite filter (after bisecting it) to control the input impedance. These formulas will be developed later in this section. Example 9.2 Design an m-derived T-section low-pass filter with a cutoff frequency of 2 MHz and a nominal impedance of 75 . Assume that f∞ = 2.05 MHz. Plot the response of this filter in the frequency band of 100 kHz to 10 MHz. SOLUTION From (9.1.31), 1 − m2 = Therefore,

m=

1−

fc f∞

fc f∞

2 =

1−

2

2 2.05

2 = 0.2195

345

IMAGE PARAMETER METHOD

TABLE 9.3

Design Relations for Composite Filters

Low-Pass

High-Pass

Constant-k T-filter

Constant-k T-filter Z0 =

L/2

L/2 C

√

L/C √ ωc = 2/ LC L = 2Z0 /ωc C = 2/Z0 ωc

Z0 = 2C

L

2C

√

L/C √ ωc = 1/2 LC L = 0.5Z0 /ωc C = 0.5/Z0 ωc

m-derived T-Section

m-derived T-Section

(Values of L and C are the same as above) fc 2 m= 1− f∞

(Values of L and C are the same as above) f∞ 2 m= 1− fc

mL/2

mL/2

2C/m

1 − m2 L 4m

L/m 4m C 1 − m2

mC

Input and Output Matching Sections

0.3L 8 L 15 0.3C

2C/m

Input and Output Matching Sections

0.3L 8L 15 0.3C

C/0.3 L/0.3 15 C 8

C/0.3 L/0.3 15 C 8

Using (9.1.24) and (9.1.25) or Table 9.3 and component values of the constantk section obtained earlier in Example 9.1, the m-derived filter is designed as follows: mL/2 = 0.2195 × 5.9683 = 1.31 µH mC = 0.2195 × 2.122 = 465.78 nF and

1 − m2 L = 12.94 µH 4m

This filter circuit is illustrated in Figure 9.8.

346

FILTER DESIGN 1.31 µH

1.31 µH

12.94 µH 465.78 pF

Figure 9.8

An m-derived T-section for Example 9.2.

As noted earlier, the image impedance of this circuit will be the same as that of the corresponding constant-k section. Hence, it will vary with frequency as illustrated in Figure 9.4. The propagation constant γ of this circuit is determined from the formula listed in Table 9.1. The transfer characteristics are then found as e−γ . Its magnitude and phase characteristics versus frequency are illustrated in Figure 9.9. The transfer characteristics illustrated in Figure 9.9 indicate that the m-derived filter has a sharp change at a cutoff frequency of 2 MHz. However, the output signal rises to −4 dB in its stopband. On the other hand, the constant-k filter provides higher attenuation in its stopband. For example, the m-derived filter characteristic in Figure 9.9 shows only 4-dB attenuation at 6 MHz, whereas the corresponding constant-k T-section has an attenuation of more than 30 dB at this frequency, as depicted in Figure 9.3. Composite Filters As demonstrated through the preceding examples, the m-derived filter provides an infinitely sharp attenuation right at its cutoff. However, the attenuation in its stopband is unacceptably low. Contrary to this, the constant-k filter shows a higher attenuation in its stopband, although the change is unacceptably gradual at the transition from passband to stopband. One way to solve the problem is to cascade these two filters. Since image impedance stays the same in two cases, this cascading will not create a new impedance-matching problem. However, we still need to address the problem of image impedance variation with frequency at the input and output ports of the network. As illustrated in Figure 9.7, the image impedance of the π-section with m = 0.6 remains almost constant over 90% of the passband. If this network is bisected to connect on either side of the cascaded constant-k and m-derived sections, it should provide the desired impedance characteristics. To verify these characteristics, let us consider a bisected π-section as shown in Figure 9.10. Its transmission parameters can be found easily following the procedure described in Chapter 8. From (8.4.4) to (8.4.7), we find that A=1+

Z1 4Z2

(9.1.34)

347

IMAGE PARAMETER METHOD 0.1

0.2

0.5

1

2

5

10

0.2

0.5

1

2

5

10

2

5

10

5

10

0

Magnitude (dB)

−2 −4 −6 −8 −10 0.1

Frequency (MHz) 0.1

0.2

0.5

1

0.1

0.2

0.5

0 −25

Phase (deg)

−50 −75 −100 −125 −150 −175 1

2

Frequency (MHz)

Figure 9.9 Frequency response of the m-derived T-section shown in Figure 9.8.

Z′1/2

Zi1

Figure 9.10

2Z′2

Zi2

Right-hand side of the bisected π-section shown in Figure 9.6.

348

FILTER DESIGN

Z1 2 1 C= 2Z2

B=

(9.1.35) (9.1.36)

and D=1

(9.1.37)

Image impedance Zi1 and Zi2 can now be found from (9.1.10) and (9.1.11) as follows: Z2 (9.1.38) Zi1 = Z1 Z2 + 1 = ZiT 4 and

Zi2 =

Z1 Z2 Z1 Z2 = Ziπ = 1 + Z1 /4Z2 ZiT

(9.1.39)

Thus, the bisected π-section can be connected at the input and output ports of cascaded constant-k and m-derived sections to obtain a composite filter that solves the impedance problem as well. Relations for the design of these circuits are summarized in Table 9.3. Example 9.3 Design a low-pass composite filter with a cutoff frequency of 2 MHz and an image impedance of 75 in its passband. Assume that f∞ = 2.05 MHz. Plot its response in the frequency range 100 kHz to 10 MHz. SOLUTION Constant-k and m-derived sections of this filter are designed in Examples 9.1 and 9.2, respectively. The filter’s input and output matching sections can be designed as follows. With m = 0.6, we find that mL = 3.581 µH 2 mC = 636.6 pF 2 and 1 − m2 L = 6.366 µH 2m This composite filter is depicted in Figure 9.11. Figure 9.12 illustrates the frequency response of this composite filter. As it indicates, there is a fairly sharp change in output signal as the frequency changes from its passband to the stopband. At the same time, the output stays well below −40 dB in its stopband. Figure 9.13 shows variation in the image impedance

349

IMAGE PARAMETER METHOD

1.31 µH

3.581 µH 6.366 µH

1.31 µH

5.9683 µH 5.9683 µH

3.581 µH

12.94 µH

6.366 µH

465.78 pF

636.6 pF

2.122 nF 636.6 pF

Bisected Π

Constant-k

Bisected Π

m-derived

Figure 9.11 Composite filter of Example 9.3. 0.1

0.2

0.5

1

2

5

10

0.2

0.5

1

2

5

10

2

5

10

0

Magnitude (dB)

−10 −20 −30 −40 −50 0.1

Frequency (MHz) 0.1

0.2

0.5

1

0.2

0.5

1

150

Phase (deg)

100 50 0 −50 −100 −150 0.1

2

5

10

Frequency (MHz)

Figure 9.12 Frequency response of the composite filter of Figure 9.11.

350

FILTER DESIGN

200

Magnitude (Ω)

150

100

50

1

2 Frequency (MHz)

3

4

75

Phase (deg)

50 25 1 −25

2

3

4

Frequency (MHz)

−50 −75

Figure 9.13 Image impedance of the composite filter of Figure 9.11 versus frequency.

of this filter as the signal frequency changes. This indicates that the image impedance stays at 75 (pure real, because the phase angle is zero) over most of its passband. Example 9.4 Design a high-pass composite filter with a nominal impedance of 75 . It must pass all signals over 2 MHz. Assume that f∞ = 1.95 MHz. Plot its characteristics in the frequency range 1 to 10 MHz. SOLUTION From Table 9.3, we find the components of its constant-k section as follows: L=

75 H = 2.984 µH 2 × 2 × π × 2 × 106

C=

1 F = 530.5 pF 2 × 2 × π × 2 × 106 × 75

and

351

IMAGE PARAMETER METHOD

Similarly, the component values for its m-derived filter section are determined as follows: f∞ 2 1.95 2 m= 1− = 1− = 0.222 fc 2 Hence, 2C = 4.775 nF m L = 13.43 µH m and 4m C = 0.496 nF 1 − m2 The component values for the bisected π-section to be used at its input and output ports are found as C = 1.768 nF 0.3 L = 9.947 µH 0.3 and 15 C = 0.9947 nF 8 The composite filter (after simplifying for the series capacitors in various sections) is illustrated in Figure 9.14. The frequency response of this composite filter is depicted in Figure 9.15. It shows that the attenuation in its stopband stays below −40 dB, and the switching to passband is fairly sharp. As usual with this type of circuit, its phase characteristics may not be acceptable for certain applications because of inherent distortion. 0.6631 nF

9.947 µH

0.9947 nF

0.8681 nF

2.984 µH

1.2903 nF

13.43 µH

0.496 nF

9.947 µH

0.9947 nF

Figure 9.14 Composite high-pass filter with 2-MHz cutoff frequency.

352

FILTER DESIGN 1

1.5

2

1

1.5

2

3

5

7

10

3 5 Frequency (MHz)

7

10

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

1

1.5

2

3

1

1.5

2

5

7

3 5 Frequency (MHz)

7

10

150

Phase (deg)

100 50 0 −50 −100 −150 10

Figure 9.15 Frequency response of the composite high-pass filter of Figure 9.14.

Magnitude (Ω)

250 200 150 100 50 5 10 15 Frequency (MHz)

20

5 10 15 Frequency (MHz)

20

Phase (deg)

80 60 40 20

Figure 9.16 Image impedance of the composite high-pass filter of Figure 9.14 as a function of signal frequency.

INSERTION-LOSS METHOD

353

As depicted in Figure 9.16, the image impedance of this composite filter is almost constant at 75 . Note that its magnitude varies over a wide range in the stopband while its phase angle seems to remain constant at 90◦ . The phase angle of the image impedance is zero in the passband.

9.2 INSERTION-LOSS METHOD The output of an ideal filter would be the same as its input in the passband, whereas it would be zero in the stopband. The phase response of this filter must be linear to avoid signal distortion. In reality, such circuits do not exist and a compromise is needed to design the filters. The image parameter method described in Section 9.1 provides a simple design procedure. However, the transfer characteristics of this circuit cannot be shaped as desired. On the other hand, the insertion-loss method provides ways to shape pass- and stopbands of the filter, although its design theory is much more complex. The power-loss ratio of a two-port network is defined as the ratio of the power that is delivered to the load when it is connected directly at the generator to the power delivered when the network is inserted between the two. In other words, power incident at port 1 1 = power delivered to the load connected at port 2 1 − |(ω)|2 (9.2.1) The power-loss ratio, PLR , expressed in decibels, is generally known as the insertion loss of the network. It can be proved that |(ω)|2 must be an even function of ω for a physically realizable network. Therefore, polynomials of ω2 can represent it as follows. PLR =

|(ω)|2 =

f1 (ω2 ) f1 (ω2 ) + f2 (ω2 )

(9.2.2)

f1 (ω2 ) f2 (ω2 )

(9.2.3)

and PLR = 1 +

where f1 (ω2 ) and f2 (ω2 ) are real polynomials in ω2 . Alternatively, the magnitude of the voltage gain of the two-port network can be found as 1 1 |G(ω)| = √ = (9.2.4) PLR 1 + f1 (ω2 )/f2 (ω2 ) Hence, if PLR is specified, (ω) is fixed. Therefore, the insertion-loss method is similar to the impedance-matching methods discussed in Chapter 7. Traditionally, the filter design begins with a lumped-element low-pass network that is synthesized by using normalized tables. It is subsequently scaled to the desired cutoff frequency and the impedance. Also, the low-pass prototype can

354

FILTER DESIGN

G(ω)

1

0

Passband

Stopband ωc

ω

Figure 9.17 Characteristics of an ideal low-pass filter.

be transformed to obtain a high-pass, a bandpass, or a bandstop filter. These lumped-element filters are used as a starting point to design the transmission line filter. In this section we present the design procedure for two different types of lumped-element low-pass filters. It is followed by the transformation techniques used to design high-pass, bandpass, and bandstop filters. Low-Pass Filters As illustrated in Figure 9.17, an ideal low-pass filter will pass the signals below its cutoff frequency ωc without attenuation while it will stop those with higher frequencies. Further, the transition from its passband to its stopband will be sharp. In reality, this type of filter cannot be designed. Several approximations to these characteristics are available that can be physically synthesized. Two of these are presented below. Maximally Flat Filter As its name suggests, this type of filter provides the flattest possible passband response. However, its transition from passband to stopband is gradual. This is also known as a binomial or Butterworth filter. The magnitude of its voltage gain (and hence, its frequency response) is given as follows: 1 |G(ω)| = 1 + ζω2n

n = 1, 2, 3, . . .

(9.2.5)

where ζ is a constant that controls the power-loss ratio at its band edge, n is the order, and ω is the normalized frequency. For |G(ω)| = 0.7071 (i.e., −3 dB) at the band edge, ζ is unity. Note that derivatives of |G(ω)| at normalized frequencies much below the band edge are close to zero. This guarantees maximum flatness in the passband. Further, (9.2.5) indicates that |G(ω)| will be a fractional number for ω greater than zero. Therefore, it is generally known as the insertion loss, L, of the network when expressed in decibels. Hence, 2n 1 ω = 10 log10 1 + ζ dB L = −20 log10 2n ω c 1 + ζ (ω/ωc ) (9.2.6)

355

INSERTION-LOSS METHOD

Normalized frequency 0.2

0.4

0.6

0.8

−0.5

1 n=2

−1 G(ω)(dB)

1.2

n=4

−1.5

n=7

−2 −2.5

Figure 9.18 Characteristics of maximally flat low-pass filters.

2n ω L = log10 1 + ζ 10 ωc

or

or

ω ζ ωc

2n = 10L/10 − 1

(9.2.7)

Figure 9.18 illustrates the passband characteristics of this type of filter for three different values of n. The power-loss ratio at its band edge is assumed to be −3 dB, and therefore ζ is equal to unity. It shows that the passband becomes flatter with a sharper cutoff when its order n is increased. However, this relationship is not a linear one. This change in characteristics is significant for lower values of n. Equation (9.2.7) can be used for determining the ζ and n values of a filter as follows. If the insertion loss at the band edge (ω = ωc ) is specified as L = Lc , (9.2.7) yields ζ = 10Lc /10 − 1 (9.2.8) Similarly, the required order n (and hence, number of elements required) of a filter can be determined for the L specified at a given stopband frequency. It is found as follows: n=

1 log10 (10L/10 − 1) − log10 ζ × 2 log10 (ω/ωc )

(9.2.9)

Chebyshev Filter A filter with a sharper cutoff can be realized at the cost of flatness in its passband. Chebyshev filters possess ripples in the passband but provide a sharp transition

356

FILTER DESIGN

into the stopband. In this case, Chebyshev polynomials are used to represent the insertion loss. Mathematically, 1 |G(ω)| = 1 + ζTm2 (ω)

m = 1, 2, 3, . . .

(9.2.10)

where ζ is a constant, ω is normalized frequency, and Tm (ω) is a Chebyshev polynomial of the first kind and degree m, defined in (7.5.2). Figure 9.19 shows the frequency response of a typical Chebyshev filter for m = 7. It assumes that ripples up to −3 dB in its passband are acceptable. A comparison of this characteristic to that for a seventh-order Butterworth filter shown in Figure 9.18 indicates that it has a much sharper transition from passband to stopband. However, it is achieved at the cost of ripples in its passband. As before, the insertion loss of a Chebyshev filter is found as follows:

1

L = −20 log10 1 + ζTm2 (ω/ωc ) or

= 10 log10 1 +

2 −1 ω 1 + ζ cos m cos 10 log 10 ωc L= ω 10 log10 1 + ζ cosh2 m cosh−1 ωc

ζTm2

ω ωc

0 ≤ ω ≤ ωc (9.2.11) ωc < ω

where ζ = 100.1×Gr − 1

(9.2.12)

Normalized frequency 0.2

0.4

0.6

0.8

1

1.2

−1

G(ω) (dB)

−2 −3 −4 −5 −6 −7

Figure 9.19

Characteristics of a low-pass Chebyshev filter for m = 7.

357

INSERTION-LOSS METHOD

Gr is the ripple amplitude in decibels. The order m (and hence, number of elements) of a Chebyshev filter can be found from its characteristics as follows: m=

cosh−1

(100.1×L − 1)/(100.1×Gr − 1) cosh−1 (ω/ωc )

(9.2.13)

where L is required insertion loss in decibels at a specified frequency ω. Example 9.5 It is desired to design a maximally flat low-pass filter with at least 15 dB attenuation at ω = 1.3 ωc and −3 dB at its band edge. How many elements will be required for this filter? If a Chebyshev filter is used with a 3-dB ripple in its passband, find the number of circuit elements. SOLUTION From (9.2.8), ζ = 10Lc /10 − 1 = 100.3 − 1 = 1 and from (9.2.9) we have n=

1 log10 (10L/10 − 1) − log10 ζ log10 (101.5 − 1) × = 0.5 × = 6.52 2 log10 (ω/ωc ) log10 (1.3)

Therefore, seven elements will be needed for this maximally flat filter. In the case of a Chebyshev filter, (9.2.13) gives m=

cosh−1

√ cosh−1 101.5 − 1 (100.1×L − 1)/(100.1×Gr − 1) = = 3.17 cosh−1 (ω/ωc ) cosh−1 (1.3)

Hence, it will require only three elements. The characteristics of these two filters are illustrated in Figure 9.20.

Normalized frequency

−1

0.5

1

1.5

2

G(ω) (dB)

−2 −3 −4 −5 −6 −7

Figure 9.20 Characteristics of low-pass Butterworth and Chebyshev filters.

358

FILTER DESIGN

Low-Pass Filter Synthesis The scope of this book does not include the theory of passive filter synthesis. Several excellent references are available in the literature for those who may be interested in this. The design procedure presented here is based on a doubly terminated low-pass ladder network, as shown in Figure 9.21. The g notation used in the figure signifies the roots of an nth-order transfer function that govern its characteristics. These represent the normalized reactance values of filter elements with a cutoff frequency ωc = 1. The source resistance is represented by g0 , and the load is gn+1 . The filter is made up of series inductors and shunt capacitors that are in the form of cascaded T-networks. Another possible configuration is the cascaded π-network that is obtained after replacing g1 by a short circuit, connecting a capacitor across the load gn+1 , and renumbering filter elements 1 through n. Elements of this filter are determined from the n roots of the transfer function. The transfer function is selected according to the pass- and stopband characteristics desired. Normalized values of elements are then found from the roots of that transfer function. These values are then adjusted according to the desired cutoff frequency and the source and load resistance. Design procedures for the maximally flat and the Chebyshev filters can be summarized as follows. Assume that the cutoff frequency is given as follows: ωc = 1

(9.2.14)

Butterworth and Chebyshev filters can then be designed using the following formulas. For a Butterworth filter, g0 = gn+1 = 1

(9.2.15)

and gp = 2 sin

(2p − 1)π 2n

p = 1, 2, · · · , n

(9.2.16)

Element values computed from (9.2.15) and (9.2.16) are given in Table 9.4 for n up to 7.

g1

g0

g3

g2

g5

g4

gn−2

gn

gn−1

Figure 9.21 Low-pass ladder network prototype.

gn+1

359

INSERTION-LOSS METHOD

TABLE 9.4 ωc = 1)

Element Values for Low-Pass Binomial Filter Prototypes (g0 = 1,

n

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

2.0000 1.4142 1.0000 0.7654 0.6180 0.5176 0.4450

1.0000 1.4142 2.0000 1.8478 1.6180 1.4142 1.2470

1 1.0000 1.8478 2.0000 1.9319 1.8019

1.0000 0.7654 1.6180 1.9319 2.0000

1.0000 0.6180 1.4142 1.8019

1 0.5176 1.2470

1 0.445

1.0000

For a Chebyshev filter, g0 = 1 1 gm+1 = ξ coth 4

(9.2.17) m is an odd number m is an even number

g1 =

2a1 χ

gp =

4a(p−1) ap b(p−1) g(p−1)

(9.2.18)

(9.2.19)

and p = 2, 3, · · · , m

(9.2.20)

where

Gr ξ = ln coth 17.37 χ = sinh ap = sin

ξ 2m

(9.2.21) (9.2.22)

(2p − 1)π 2m

(9.2.23)

pπ m

(9.2.24)

and bp = χ2 + sin2

The element values for several low-pass Chebyshev filters that are computed from (9.2.17) to (9.2.24) are given in Tables 9.5 to 9.9.

360

FILTER DESIGN

TABLE 9.5 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 0.1 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

0.3053 0.8431 1.0316 1.1088 1.1468 1.1681 1.1812

1.000 0.6220 1.1474 1.3062 1.3712 1.4040 1.4228

1.3554 1.0316 1.7704 1.9750 2.0562 2.0967

1.0000 0.8181 1.3712 1.5171 1.5734

1.3554 1.1468 1.9029 2.0967

1.0000 0.8618 1.4228

1.3554 1.1812

1.0000

TABLE 9.6 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 0.5 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

0.6987 1.4029 1.5963 1.6703 1.7058 1.7254 1.7373

1.0000 0.7071 1.0967 1.1926 1.2296 1.2479 1.2582

1.9841 1.5963 2.3662 2.5409 2.6064 2.6383

1.0000 0.8419 1.2296 1.3136 1.3443

1.9841 1.7058 2.4759 2.6383

1.0000 0.8696 1.2582

1.9841 1.7373

1.0000

TABLE 9.7 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 1.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.0178 1.8220 2.0237 2.0991 2.1350 2.1547 2.1666

1.0000 0.6850 0.9941 1.0644 1.0911 1.1041 1.1115

2.6599 2.0237 2.8312 3.0010 3.0635 3.0937

1.0000 0.7892 1.0911 1.1518 1.1735

2.6599 2.1350 2.9368 3.0937

1.0000 0.8101 1.1115

2.6599 2.1666

1.0000

Scaling the Prototype to the Desired Cutoff Frequency and Load ž

Frequency scaling. For scaling the frequency from 1 to ωc , divide all normalized g values that represent capacitors or inductors by the desired cutoff frequency expressed in radians per second. Resistors are excluded from this operation.

361

INSERTION-LOSS METHOD

TABLE 9.8 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 2.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.5297 2.4883 2.7108 2.7926 2.8311 2.8522 2.8651

1.0000 0.6075 0.8326 0.8805 0.8984 0.9071 0.9120

4.0957 2.7108 3.6064 3.7829 3.8468 3.8776

1.0000 0.6818 0.8984 0.9392 0.9536

4.0957 2.8311 3.7153 3.8776

1.0000 0.6964 0.9120

4.0957 2.8651

1.0000

TABLE 9.9 Element Values for Low-Pass Chebyshev Filter Prototypes (g0 = 1, ωc = 1, and 3.0 dB ripple) m

g1

g2

g3

g4

g5

g6

g7

g8

1 2 3 4 5 6 7

1.9954 3.1014 3.3489 3.4391 3.4815 3.5047 3.5187

1.0000 0.5339 0.7117 0.7483 0.7619 0.7685 0.7722

5.8095 3.3489 4.3473 4.5378 4.6063 4.6392

1.0000 0.5920 0.7619 0.7929 0.8038

5.8095 3.4815 4.4643 4.6392

1.0000 0.6033 0.7722

5.8095 3.5187

1.0000

ž

Impedance scaling. To scale g0 and gn+1 to X from unity, multiply all g values that represent resistors or inductors by X. On other hand, divide those g values representing capacitors by X.

Example 9.6 Design a Butterworth filter with a cutoff frequency of 10 MHz and an insertion loss of 30 dB at 40 MHz. It is to be used between a 50- load and a generator with an internal resistance of 50 . SOLUTION From (9.2.9), we have n= =

1 log10 (1030/10 − 1) − log10 (103/10 − 1) × 2 log10 (40/10) 1 log10 (103 − 1) − log10 (1.9953 − 1) 0.5 × 3 × ≈ ≈ 2.5 2 log10 (4) 0.6

Since the number of elements must be an integer, selecting n = 3 will provide more than the attenuation specified at 40 MHz. From (9.2.15) and (9.2.16), g0 = g4 = 1

362

FILTER DESIGN 795.77 nH 50 Ω R0 = Rs

Figure 9.22

795.77 nH

L1

L3 R4 = RL

636.62 pF

C2

50 Ω

Maximally flat low-pass filter obtained in Example 9.6.

π =1 6 π g2 = 2 sin = 2 2 g1 = 2 sin

and g3 = 2 sin

5π =1 6

If we use the circuit arrangement illustrated in Figure 9.22, element values can be scaled to match the frequency and load resistance. Using the two rules, these values are found as follows: L1 = L3 = 50 ×

1 H = 795.77 nH 2π × 107

and C2 =

1 2 F = 636.62 pF × 50 2π × 107

The frequency response of this filter is shown in Figure 9.23. It indicates a 6dB insertion loss in the passband of the filter. This happens because the source resistance is considered a part of this circuit. In other words, this represents voltage across RL with respect to source voltage, not to input of the circuit. Since source resistance is equal to the load, there is equal division of source voltage that results in −6 dB. Another 3-dB loss at the band edge shows a total of about 9 dB at 10 MHz. This characteristic shows that there is an insertion loss of over 40 dB at 40 MHz. It is clearly more than 30 dB from the band edge, as required by the example. Also, there is a nonlinear phase variation in its passband. Alternatively, we can use a circuit topology as shown in Figure 9.24. In that case, component values are found by using the scaling rules: L2 = 50 ×

1 × 2 = 0.15915 × 10−5 H = 1591.5 nH 2π × 107

C1 = C3 =

1 1 10−9 × F = 318.31 pF ×1= 7 50 2π × 10 π

and

363

INSERTION-LOSS METHOD 0.1

0.5

1

0.1

0.5

1

0.1

0.5

1

0.1

0.5

1

5

10

50

100

5 10 Frequency (MHz)

50

100

10

50

100

5 10 Frequency (MHz)

50

100

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

5

150

Phase (deg)

100 50 0 −50 −100 −150

Figure 9.23 Characteristics of the low-pass filter shown in Figure 9.22.

50 Ω R1 = Rs

C1

1.5915 µH L2

318.31 pF

Figure 9.24

R4 = RL

C3 318.31 pF

50 Ω

Alternative circuit for Example 9.6.

The frequency response of this circuit is illustrated in Figure 9.25. It is identical to that shown in Figure 9.23 for the earlier circuit. Example 9.7 Design a low-pass Chebyshev filter that may have ripples no more than 0.01 dB in its passband. The filter must pass all frequencies up to 100 MHz

364

FILTER DESIGN 0.1

0.5

1

0.1

0.5

1

5

10

50

100

5

10

50

100

5

10

50

100

5

10

50

100

0

Magnitude (dB)

−10 −20 −30 −40 −50 −60

Frequency (MHz) 0.1

0.5

1

0.1

0.5

1

150

Phase (deg)

100 50 0 −50 −100 −150 Frequency (MHz)

Figure 9.25 Characteristics of the low-pass filter shown in Figure 9.24.

and attenuate the signal at 400 MHz by at least 5 dB. The load and the source resistance are of 75 each. SOLUTION Since Gr = 0.01 dB and

400 L 100

= 5 dB

INSERTION-LOSS METHOD

365

(9.2.13) gives m=

cosh−1

(100.5 − 1)/(100.001 − 1) =2 cosh−1 (4)

Since we want a symmetrical filter with 75 on each side, we select m = 3 (an odd number). It will provide more than 5 dB of insertion loss at 400 MHz. The g values can now be determined from (9.2.17) to (9.2.24), as follows. From (9.2.23), π = 0.5 6 π a2 = sin = 1 2 a1 = sin

and a3 = sin

5π = 0.5 6

Next, from (9.2.21), (9.2.22), and (9.2.24), we get 0.01 ξ = ln coth = 7.5 17.37 7.5 = 1.6019 6 π b1 = 1.60192 + sin2 = 3.316 3 2π b2 = 1.60192 + sin2 = 3.316 3 χ = sinh

and b3 = 1.60192 + sin2

3π = 2.566 3

However, b3 is not needed in further calculations. Finally, from (9.2.17) and (9.2.20), g0 = g4 = 1 2 × 0.5 = 0.62425 1.6019 4 × 0.5 × 1 g2 = = 0.9662 3.316 × 0.62425 g1 =

and g3 =

4 × 1 × 0.5 = 0.62425 3.316 × 0.9662

366

FILTER DESIGN

74.5155 nH 75 Ω R0 = R s

Figure 9.26

74.5155 nH L3

L1 20.5 pF

R4 = R L C2

75 Ω

Low-pass Chebyshev filter circuit obtained in Example 9.7.

For the circuit topology of Figure 9.26, element values are found after applying the scaling rules as follows: L1 = L3 =

75 × 0.62425 H = 74.5155 nH 2π × 108

and C2 =

1 1 × × 0.9662 F = 20.5 pF 75 2π × 108

The frequency response of this filter is illustrated in Figure 9.27. As expected, there is over a 20-dB insertion loss at 400 MHz compared with its passband. Note that the magnitude of ripple allowed is so small that it does not show up in this figure. However, it is present there, as shown on an expanded scale in Figure 9.28. Figure 9.28 shows the passband characteristics of this Chebyshev filter. Since the scale-of-magnitude plot is now expanded, the passband ripple is clearly visible. As expected, the ripple stays between −6.02 and −6.03 dB. Phase variation in this passband ranges from 0◦ to −70◦ . The band edge of this filter can be sharpened further either by using a higher-order filter or by allowing larger magnitudes of the ripple in its passband. The higher-order filter will require more elements because the two are directly related. The next example illustrates that a sharper transition between the bands can be obtained even with a three-element filter if ripple magnitudes up to 3 dB are acceptable. Example 9.8 Reconsider Example 9.7 to design a low-pass filter that exhibits the Chebyshev response with 3-dB ripple in its passband, m = 3, and a cutoff frequency of 100 MHz. The filter must have 75 at both its input and output ports. SOLUTION From (8.2.23), π = 0.5 6 π a2 = sin = 1 2

a1 = sin

367

INSERTION-LOSS METHOD 1

5

10

1

5

10

50

100

500 1000

50

100

500 1000

0

Magnitude (dB)

−10 −20 −30 −40 −50 Frequency (MHz) 1

5

10

50

100

500 1000

1

5

10

50

100

500 1000

150 100

Phase (deg)

50 0 −50 −100 −150

Frequency (MHz)

Figure 9.27 Characteristics of the low-pass Chebyshev filter shown in Figure 9.26.

and a3 = sin

5π = 0.5 6

Now, from (9.2.21), (9.2.22), and (9.2.24),

3 ξ = ln coth 17.37 χ = sinh

= 1.7661

1.7661 = 0.2986 6

368

FILTER DESIGN

1

2

5

1

2

5

10

20

50

100

10

20

50

100

Magnitude (dB)

−6.022 −6.024 −6.026 −6.028 −6.03 Frequency (MHz) 1

2

5

1

2

5

10

20

50

100

10

20

50

100

0 −10

Phase (deg)

−20 −30 −40 −50 −60

Frequency (MHz)

Figure 9.28 Passband characteristics of the filter shown in Figure 9.26.

π = 0.8392 3 2π b2 = 0.29862 + sin2 = 0.8392 3

b1 = 0.29862 + sin2

and b3 = 0.29862 + sin2 However, b3 is not needed.

3π = 0.0892 3

369

INSERTION-LOSS METHOD 0.4 µH 75 Ω

L1

R0 = Rs

Figure 9.29

0.4 µH L3 R4 = RL

15.1 pF

C2

75 Ω

Low-pass Chebyshev filter circuit obtained in Example 9.8.

The g values for the filter can now be determined from (9.2.17) to (9.2.20), as follows: g0 = g4 = 1 2 × 0.5 = 3.349 0.2986 4 × 0.5 × 1 g2 = = 0.7116 3.349 × 0.8392

g1 =

and g3 =

4 × 1 × 0.5 = 3.349 0.7116 × 0.8392

These g values are also listed in Table 9.9 for m = 3. Elements of the circuit shown in Figure 9.29 are determined using the scaling rules, as follows (see Figure 9.30 for the characteristics): L1 = L3 = and C2 =

75 × 3.349 H = 0.4 µH 2π × 108

1 1 × 0.7116 F = 15.1 pF × 75 2π × 108

High-Pass Filter As mentioned earlier, a high-pass filter can be designed by transforming the low-pass prototype. This frequency transformation is illustrated in Figure 9.31. As illustrated, an ideal low-pass filter passes all signals up to the normalized frequency of unity with zero insertion loss, whereas it attenuates higher frequencies completely. On the other hand, a high-pass filter must pass all signals with frequencies higher than its cutoff frequency ωc and stop the signals that have lower frequencies. Therefore, the following frequency transformation formula will transform a low-pass filter to a high-pass filter: ω=−

ωc ω

(9.2.25)

370

FILTER DESIGN 1

10

100

1

10 Frequency (MHz)

100

1

10

100

1

10 Frequency (MHz)

100

−6 −7 Magnitude (dB)

−8 −9 −10 −11 −12 −13

100

Phase (deg)

50 0 −50 −100 −150

Characteristics of the low-pass filter of Example 9.8.

Magnitude

Magnitude

Figure 9.30

1

0

1

ω

1

0

ωc

Figure 9.31 Transformation from a low-pass to a high-pass characteristic.

ω

INSERTION-LOSS METHOD

371

Thus, inductors and capacitors will change their places. Inductors will replace the shunt capacitors of the low-pass filter and capacitors will be connected in series, in place of inductors. These elements are determined as follows: CHP =

1 ωc gL

(9.2.26)

LHP =

1 ωc gC

(9.2.27)

and

Capacitor CHP and inductor LHP can now be scaled as required by the load and source resistance. Example 9.9 Design a high-pass Chebyshev filter with passband ripple magnitude less than 0.01 dB. It must pass all frequencies over 100 MHz and exhibit at least 5 dB of attenuation at 25 MHz. Assume that the load and source resistances are at 75 each. SOLUTION The low-pass filter designed in Example 9.7 provides the initial data for this high-pass filter. With m = 3, gL = 0.62425, and gC = 0.9662, we find from (9.2.26) and (9.2.27) that CHP =

1 F = 2.5495 nF 2π × 108 × 0.62425

and LHP =

2π ×

1 H = 1.6472 nH × 0.9662

108

Now, applying the resistance scaling, we get C1 = C3 =

2.5495 nF = 33.9933 pF = 34 pF 75

and L2 = 75 × 1.6472 = 123.5 nH The resulting high-pass Chebyshev filter is shown in Figure 9.32. Its frequency response is illustrated in Figure 9.33. As before, the source resistance is considered a part of the filter circuit and therefore its passband shows a 6-dB insertion loss. Bandpass Filter A bandpass filter can be designed by transforming the low-pass prototype as illustrated in Figure 9.34. Here, an ideal low-pass filter passes all signals up to

372

FILTER DESIGN

34 pF

75 Ω R0 = Rs

34 pF

C1 L2

Figure 9.32

R4 = RL

C3 123.5 nH

75 Ω

High-pass Chebyshev filter obtained for Example 9.8.

20

50

100

200

500

1000

2000

20

50

100

200

500

1000

2000

−7.5

Magnitude (dB)

−10 −12.5 −15 −17.5 −20 −22.5

Frequency (MHz) 20

50

100

200

500

1000

2000

20

50

100

200

500

1000

2000

150

Phase (deg)

100 50 0 −50 −100 −150

Frequency (MHz)

Figure 9.33

Characteristics of the high-pass Chebyshev filter shown in Figure 9.32.

373

Magnitude

Magnitude

INSERTION-LOSS METHOD

1

−1

0

1

ω

1

0

ωl

ω

ωu

Figure 9.34 Transformation from a low-pass to a bandpass characteristic.

the normalized frequency of unity with zero insertion loss, whereas it attenuates higher frequencies completely. On the other hand, a bandpass filter must pass all signals with frequencies between ωl and ωu and stop the signals that are outside this frequency band. Hence, the following frequency relation transforms the response of a low-pass filter to a bandpass: ω=

ω2 − ω20 1 ωu − ωl ω

where ω0 =

√

ωl × ωu

(9.2.28)

(9.2.29)

This transformation replaces the series inductor of low-pass prototype with an inductor LBP1 and a capacitor CBP1 that are connected in series. The component values are determined as follows: CBP1 =

ωu − ωl ω20 gL

(9.2.30)

LBP1 =

gL ωu − ωl

(9.2.31)

and

Also, the capacitor CBP2 that is connected in parallel with an inductor LBP2 will replace the shunt capacitor of the low-pass prototype. These elements are found as follows: LBP2 =

ωu − ωl ω20 gC

(9.2.32)

CBP2 =

gC ωu − ωl

(9.2.33)

and

These elements need to be scaled further as desired by the load and source resistance.

374

FILTER DESIGN

Example 9.10 Design a bandpass Chebyshev filter that exhibits no more than 0.01-dB ripples in its passband. It must pass signals in the frequency band 10 to 40 MHz with zero insertion loss. Assume that the load and source resistances are at 75 each. SOLUTION From (9.2.29), we have f0 =

fl fu = 107 × 40 × 106 = 20 × 106 Hz

Now, using (9.2.30) to (9.2.33) with gL = 0.62425 and gC = 0.9662 from Example 9.7, we get 2π × 106 (40 − 10) F = 19.122 nF (2π × 20 × 106 )2 × 0.62424 0.62424 = H = 3.3116 nH 2π × 30 × 106

CBP1 = LBP1

LBP2 =

2π × 106 (40 − 10) H = 12.354 nH (2π × 20 × 106 )2 × 0.9662

CBP2 =

0.9662 F = 5.1258 nF 2π × 30 × 106

and

Next, values of elements are determined after applying the load and source resistance scaling. Hence, 19.122 nF = 254.96 pF ≈ 255 pF 75 L1 = L3 = 75 × 3.3116 nH = 0.2484 µH C1 = C3 =

L2 = 75 × 12.354 nH = 0.9266 µH and C2 =

5.1258 nF = 68.344 pF 75

The resulting filter circuit is shown in Figure 9.35, and its frequency response is depicted in Figure 9.36. It illustrates that we indeed have transformed the low-pass prototype to a bandpass filter for the 10- to 40-MHz frequency band. As before, the source resistance is considered a part of the filter circuit, and therefore its passband shows a 6-dB insertion loss. Figure 9.37 shows its passband characteristics with expanded scales. It indicates that the ripple magnitude is limited to 0.01 dB, as desired.

375

INSERTION-LOSS METHOD

75 Ω

255 pF

C1

0.2484 µH

0.2484 µH

L1

L3

0.9266 µH L2

Figure 9.35

0

255 pF C3 75 Ω

68.344 pF C2

Bandpass filter circuit for Example 9.10.

0.1

1

10

100

1000

0.1

1

10 Frequency (MHz)

100

1000

0.1

1

10

100

1000

0.1

1

10 Frequency (MHz)

100

1000

Magnitude (dB)

−20 −40 −60 −80 −100 −120

150

Phase (deg)

100 50 0 −50 −100 −150

Figure 9.36

Characteristics of the bandpass filter shown in Figure 9.35.

Bandstop Filter A bandstop filter can be realized by transforming the low-pass prototype as illustrated in Figure 9.38. Here, an ideal low-pass filter passes all signals up to the normalized frequency of unity with zero insertion loss, whereas it completely

376

FILTER DESIGN 10

15

20

30

10

15

20 Frequency (MHz)

30

10

15

20

30

10

15

20 Frequency (MHz)

30

Magnitude (dB)

−6.022 −6.024 −6.026 −6.028 −6.03

60

Phase (deg)

40 20 0 −20 −40 −60

Figure 9.37 Passband characteristics of the bandpass filter shown in Figure 9.35.

attenuates higher frequencies. On the other hand, a bandstop filter must stop all signals with frequencies between ωl and ωu and pass the signals that are outside this frequency band. Hence, its characteristics are opposite to that of a bandpass filter considered earlier. The following frequency relation transforms the response of a low-pass filter to the bandstop: ω = (ωu − ωl )

ω ω2 − ω20

(9.2.34)

This transformation replaces the series inductor of low-pass prototype with an inductor LBS1 and a capacitor CBS1 that are connected in parallel. Component values are determined as follows: LBS1 =

(ωu − ωl )gL ω20

(9.2.35)

377

−1

1

0

ω

1

Magnitude

Magnitude

INSERTION-LOSS METHOD

1

0

ωl

ω

ωu

Figure 9.38 Transformation from a low-pass to a bandstop characteristic.

and CBS1 =

1 (ωu − ωl )gL

(9.2.36)

Also, capacitor CBS2 , which is connected in series with an inductor LBS2 , will replace the shunt capacitor of low-pass prototype. These elements are found as follows: LBS2 =

1 (ωu − ωl )gC

(9.2.37)

CBS2 =

(ωu − ωl )gC ω20

(9.2.38)

and

These elements need to be further scaled as desired by the load and source resistance. Table 9.10 summarizes these transformations for the low-pass prototype filter. Example 9.11 Design a maximally flat bandstop filter with n = 3. It must stop signals in the frequency range 10 to 40 MHz and pass the rest of the frequencies. Assume that the load and source resistances are at 75 each. SOLUTION From (9.2.29), we have f0 = fl fu = 107 × 40 × 106 = 20 × 106 Hz Now, using (9.2.35) to (9.2.38) with gL = 1 and gC = 2 from Example 9.6, we get 2π × 106 (40 − 10) × 1 H = 11.94 nH (2π × 20 × 106 )2 1 = F = 5.305 nF 6 2π × 10 (40 − 10) × 1

LBS1 = CBS1

378

FILTER DESIGN

TABLE 9.10 Filter Transformations Filter

Circuit Elements

Low-pass gC

gL

High-pass 1 ω c gL

1 ωc gC

Bandpass ωu − ω l ω20 gL

ωu − ω l ω20 gC gL ω u − ωl

gC ωu − ωl

Bandstop

ωu − ωl gL

1 ωu − ω l gC

ω20 ωu − ω l gC ω20 1 ωu − ωl gL

LBS2 =

1 H = 2.653 nH 2π × 106 (40 − 10) × 2

CBS2 =

2π × 106 (40 − 10) × 2 F = 23.87 nF (2π × 20 × 106 )2

and

Values of required elements are determined next following the load and source resistance scaling. Hence, C1 = C3 =

5.305 nF = 70.73 pF 75

379

INSERTION-LOSS METHOD 0.8955 µH 75 Ω

0.8955 µH

L1

L3

C1 70.73 pF 318.3 pF

75 Ω L2

0.2 µH

Figure 9.39

−6

C3 70.73 pF

C2

Bandstop circuit for Example 9.11.

0.1

1

10

100

1000

0.1

1

10 Frequency (rad/s)

100

1000

0.1

1

10

100

1000

0.1

1

10 Frequency (rad/s)

100

1000

Magnitude (dB)

−6.5 −7 −7.5 −8

100

Phase (deg)

50 0 −50 −100 −150

Figure 9.40

Characteristics of the bandstop filter shown in Figure 9.39.

380

FILTER DESIGN

L1 = L3 = 75 × 11.94 nH = 0.8955 µH L2 = 75 × 2.653 nH = 0.1989 µH ≈ 0.2 µH and C2 =

23.87 nF = 318.3 pF 75

The resulting filter circuit is shown in Figure 9.39. Its frequency response depicted in Figure 9.40 indicates that we have transformed the low-pass prototype to a bandstop filter for the 10- to 40-MHz frequency band. As before, the source resistance is considered a part of the filter circuit, and therefore its passband shows a 6-dB insertion loss.

9.3 MICROWAVE FILTERS The filter circuits presented so far in this chapter use lumped elements. However, these may have practical limitations at microwave frequencies. When the signal wavelength is short, distances between the filter components need to be taken into account. Further, discrete components at such frequencies may cease to operate due to associated parasitic elements and need to be approximated with distributed components. As found in Chapter 3, transmission line stubs can be used in place of lumped elements. However, there may be certain practical problems in implementing the series stubs. This section begins with a technique to design a low-pass filter with only parallel connected lines of different characteristic impedance values. It is known as the stepped impedance (or high-Z low-Z) filter. Since this technique works mainly for low-pass filters, the procedure to transform series reactance to shunt that utilizes Kuroda’s identities is summarized next. Redundant sections of the transmission line are used to separate filter elements, and therefore this procedure is known as redundant filter synthesis. Nonredundant circuit synthesis makes use of these sections to improve the filter response as well. Other methods of designing microwave filters include coupled transmission lines and resonant cavities. Impedance transformers, discussed in Chapter 7, are essentially bandpass filters with different impedance at the twoports. Interested readers can find detailed design procedures of all these filters in the references at the end of this chapter.

Low-Pass Filter Design Using Stepped Impedance Distributed Elements Consider a lossless transmission line of Z0 , as shown in Figure 9.41(a). If it is Figure 9.41(b), these two must have the ance matrix of the transmission line was

length d and characteristic impedance equivalent to the T-network shown in same network parameters. The impeddetermined in Example 8.4 as follows:

381

MICROWAVE FILTERS

d Z0, β (a)

Z11 − Z12

Z11 − Z12 Z12

(b)

jXL 2

jXL 2

jB

(c)

Figure 9.41 Lossless transmission line (a), equivalent symmetrical T-network (b), and (c) elements of T-networks.

[Z]line =

−j Z0 cot βd −j

Z0 sin βd

Z0 sin βd −j Z0 cot βd −j

The impedance matrix of the T-network, [Z]T , may be found as Z11 Z12 [Z]T = Z12 Z11

(9.3.1)

(9.3.2)

Therefore, if this T-network represents the transmission line, we can write Z11 = −j Z0 cot βd

(9.3.3)

and Z12 = −j

Z0 sin βd

(9.3.4)

Hence, Z12 is a capacitive reactance. Circuit elements in the series arm of the T-network are found to be inductive reactance. This is found as follows: Z11 − Z12 = −j Z0

βd cos βd − 1 sin (βd/2) = j Z0 = j Z0 tan sin βd cos (βd/2) 2

(9.3.5)

382

FILTER DESIGN

Therefore, the series element of the T-network is an inductive reactance while its shunt element is capacitive, provided that βd < π/2. For βd < π/4, (9.3.4) and (9.3.5) give B≈

βd Z0

(9.3.6)

and XL ≈ Z0 βd

(9.3.7)

Some special cases are as follows: 1. For Z0 very large, (9.3.6) becomes negligible compared with (9.3.7). Therefore, the equivalent T-network (and hence the transmission line) represents an inductor that is given by L=

Z0 βd ω

(9.3.8)

2. For Z0 very small, (9.3.6) dominates over (9.3.7). Therefore, the transmission line is effectively representing a shunt capacitance in this case. It is given by βd (9.3.9) C= ωZ0 Hence, high-impedance sections (Z0 = Zh ) of the transmission line can replace the inductors, and low-impedance sections (Z0 = Zm ) can replace the capacitors of a low-pass filter. In a design, these impedance values must be selected as far apart as possible and the section lengths are determined to satisfy (9.3.8) and (9.3.9). When combined with the scaling rules, electrical lengths of the inductive and capacitive sections are found as (βd)L =

R0 L Zh

rad

(9.3.10)

(βd)C =

Zm C R0

rad

(8.3.11)

and

where L and C are normalized element values (g values) of the filter and R0 is the filter impedance. Example 9.12 Design a three-element maximally flat low-pass filter with its cutoff frequency as 1 GHz. It is to be used between a 50- load and a generator with its internal impedance at 50 . Assume that Zh = 150 and Zm = 30 .

383

MICROWAVE FILTERS L1 = 1

L3 = 1

50 Ω R0 = 1

R4 = 1

C2 = 2

50 Ω

Figure 9.42 Maximally flat low-pass filter with normalized elements’ values for Example 9.12.

68.75° 19.1°

Figure 9.43

19.1°

Maximally flat low-pass filter for Example 9.12.

SOLUTION From (9.2.15) and (9.2.16), g0 = g4 = 1 π g1 = 2 sin = 1 6 π g2 = 2 sin = 2 2 and g3 = 2 sin

5π =1 6

If we use a circuit arrangement as illustrated in Figure 9.42, element values can be scaled for the frequency and the impedance. Using the scaling rules, these values are found as follows (see Figure 9.43): θ1 = θ3 = (βd)L =

1 × 50 ◦ = 0.3333 rad = 19.1 150

and θ2 = (βd)C =

30 × 2 ◦ = 1.2 rad = 68.75 50

Filter Synthesis Using the Redundant Elements As mentioned earlier, transmission line sections can replace the lumped elements of a filter circuit. Richard’s transformation provides the tools needed in such a

384

FILTER DESIGN

design. Further, Kuroda’s identities are used to transform series elements to a shunt configuration that facilitates the design. Richard’s Transformation Richard proposed that open- and short-circuited lines could be synthesized like lumped elements through the following transformation: = tan βd = tan

ωd vp

(9.3.12)

where vp is the phase velocity of signal propagating on the line. Since the tangent function is periodic with a period of 2π, (9.3.12) is a periodic transformation. Substituting in place of ω, the reactance of the inductor L and of the capacitor C may be written as follows: j XL = j L = j L tan βd

(9.3.13)

and j XC = −j

1 1 1 = −j = −j cot βd C C tan βd C

(9.3.14)

Comparison of (9.3.13) and (9.3.14) with the special cases considered in Section 3.2 indicates that the former represents a short-circuited line with its characteristic impedance as L, while the latter is an open-circuited with Z0 as 1/C. Filter impedance is assumed to be unity. To obtain the cutoff of a low-pass filter prototype at unity frequency, the following must hold: = tan βd = 1 → βd =

π λ →d= 4 8

(9.3.15)

Note that the transformation holds only at the cutoff frequency ωc (frequency corresponding to λ), and therefore the filter response will differ from that of its prototype. Further, the response repeats every 4 ωc . These transformations are illustrated in Figure 9.44. These stubs are known as the commensurate lines because of their equal lengths. Kuroda’s Identities As mentioned earlier, Kuroda’s identities facilitate the design of distributed element filters, providing the means to transform the series stubs into shunt, or vice versa, to separate the stubs physically, and to render the characteristic impedance realizable. Figure 9.45 illustrates four identities that are useful in such a transformation of networks. These networks employ the unit element (U.E.), which is basically a λ/8-long line at the cutoff frequency ωc , with specified characteristic

385

MICROWAVE FILTERS

ω-plane

Ω-plane L

Z0 = L

Short

1 C

Open

Z0 =

C

d

Figure 9.44 Distributed inductor and capacitor obtained from Richard’s transformation.

Z1 U.E. n2Z1

U.E. Z2

1 n2Z2

(a) Z1 n2 1 Z2

U.E. Z2

U.E. Z1

n2 (b) 1 n2Z2

1 Z2 U.E. n2Z1

U.E. Z1

n2 :1 (c) U.E. Z1

U.E. Z1

Z1

Z2

n2

n2 (d)

n2 = 1 +

1:n2 Z2 Z1

Figure 9.45 Kuroda’s identities.

impedance. Inductors represent short-circuited stubs and capacitors open-circuited stubs. Obviously, its equivalent in lumped circuit theory does not exist. The first identity is proved below. A similar procedure can be used to verify others. The circuits associated with the first identity can be redrawn as shown in Figure 9.46. A short-circuited stub of characteristic impedance Z1 replaces the

386

FILTER DESIGN Short

d

d Z1

Open

d n2Z2

n2Z1

Z2

d

Figure 9.46 Illustration of Kuroda’s first identity, shown in Figure 9.45.

series inductor that is followed by a unit element transmission line of characteristic impedance Z2 . Its transformation gives a unit element transmission line of characteristic impedance n2 Z1 and an open-circuited stub of characteristic impedance n2 Z2 in place of a shunt capacitor. Following Example 8.10, the transmission matrix of a short-circuited series stub with input impedance ZS can be found as follows. The impedance ZS of the series stub is given as ZS = j Z1 tan βd = j Z1 1 j Z1 A B = C D S 0 1

(9.3.16) (9.3.17)

Following Example 8.13, the transmission matrix of the unit element is found to be 1 j Z2 tan βd cos βd j Z2 sin βd A B = cos βd = j sin βd j tan βd C D U.E. cos βd 1 Z2 Z2 or 1 j Z2 1 A B j (9.3.18) =√ C D U.E. 1 + 2 1 Z2 Since these two elements are connected in cascade, the transmission matrix for the first circuit in Figure 9.46 is found from (9.3.17) and (9.3.18) as follows:

A C

B D

=

a

A C

B D

· S

A C

B D

= √ U.E.

1 1+

2

2 Z1 Z2 j Z2

1−

j (Z1 + Z2 )

1 (9.3.19)

387

MICROWAVE FILTERS

Now for the second circuit shown in Figure 9.46, the transmission matrix for the unit element is

A C

B D

U.E.

1

= cos βd j tan βd

j n2 Z1 tan βd 1

n2 Z 1

= √

1 1 + 2

1

j n2 Z 1

j n2 Z1 1

(9.3.20)

Since the impedance ZP of the shunt stub is given as Zp = −j n2 Z2 cot βd = −j

n2 Z2

(9.3.21)

its transmission matrix may be found, by following Example 8.11, as 1 0 A B = j C D p 1 n2 Z2

(9.3.22)

Hence, the transmission matrix of the circuit shown on right-hand side is found as 2 j n Z 1 1 = √ 1 1 + 2 j 1 1 + 2 n Z1 Z2

A C

B D

=

b

A C

B D

· U.E.

A C

B D

1−

p

2 Z1 Z2

(9.3.23)

On substituting for n2 , it is easy to show that this matrix is equal to that found in (9.3.19). Example 9.13 Design a three-element maximally flat low-pass filter with its cutoff frequency as 1 GHz. It is to be used between a 50- load and a generator with its internal impedance at 50 . SOLUTION A step-by-step procedure to design this filter is as follows: ž Step 1. As before, g values for the filter are found from (9.2.15) and (9.2.16) as

g0 = g4 = 1 π g1 = 2 sin = 1 6 π g2 = 2 sin = 2 2 and g3 = 2 sin

5π =1 6

388

FILTER DESIGN L1 = 1

L3 = 1

50 Ω R0 = 1

R4 = 1

C2 = 2

50 Ω (a)

L1 Z0 = 1

L1 Z0 = 1

L3 Z0 = 1

d

L3 Z0 = 1

d

U.E.

U.E.

Z0 = 1 d

d

C2 Z0 = 0.5 (b)

Z0 = 2

Z0 = 2

Z0 = 0.5 (d)

Z0 = 50 Ω

d U.E.

U.E. Z0 = 2

d

C2 Z0 = 0.5

d

(c)

d d

Z0 = 1

Z0 = 100 Ω Z0 = 100 Ω Z0 = 50 Ω

Z0 = 100 Ω

Z0 = 100 Ω

Z0 = 2 d = λ/8 at ω = 1

Z0 = 25 Ω (e)

Figure 9.47 Maximally flat low-pass filter for Example 9.13.

The circuit is drawn with normalized values as shown in Figure 9.47(a). ž Step 2. Using Richard’s transformation, the filter elements are replaced by stubs. Hence, short-circuited series stubs replace inductors L1 and L3 , while an open-circuited shunt stub replaces the capacitor C2 . This is shown in Figure 9.47(b). ž Step 3. Since the circuit found in step 2 cannot be fabricated in its present form, unit elements are added on its two sides. This is shown in Figure 9.47(c). ž Step 4. Using Kuroda’s identity shown in Figure 9.45(a), series stub and unit element combinations are transformed to an open-circuited shunt stub and the unit element. Since Z1 = Z2 = 1, equivalent capacitance C and the characteristic impedance of the corresponding stub are found to be C = 12 = 0.5 and Z0 = 2. Also, the characteristic impedance of each unit element is found to be 2. This circuit is illustrated in Figure 9.47(d). ž Step 5. For impedance and frequency scaling, all normalized characteristic impedances are now multiplied by 50 , and the stub lengths are selected as λ/8 at 1 GHz. This is shown in Figure 9.47(e).

PROBLEMS

389

SUGGESTED READING Bahl, I., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Davis, W. A., Microwave Semiconductor Circuit Design. New York: Van Nostrand Reinhold, 1984. Elliott, R. S., An Introduction to Guided Waves and Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1993. Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Matthaei, G. L., L. Young, and E. M. T. Jones, Microwave Filters, Impedance-Matching Networks, and Coupling Structures. Dedham, MA: Artech House, 1980. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998.

PROBLEMS 9.1. Design a low-pass composite filter with nominal impedance of 50 . It must pass all signals up to 10 MHz. Assume that f∞ = 10.05 MHz. Plot its characteristics in the frequency range 1 to 100 MHz. 9.2. Design a high-pass T-section constant-k filter that has a nominal impedance of 50 and a cutoff frequency of 10 MHz. Plot its frequency response in the frequency band 100 kHz to 100 MHz. 9.3. Design an m-derived T-section high-pass filter with a cutoff frequency of 10 MHz and nominal impedance of 50 . Assume that f∞ = 9.95 MHz. Plot the response of this filter in the frequency band 100 kHz to 100 MHz. 9.4. Design a high-pass composite filter with a cutoff frequency of 10 MHz and image impedance of 50 in its passband. Assume that f∞ = 9.95 MHz. Plot its response in the frequency range 100 kHz to 100 MHz. 9.5. It is desired to design a maximally flat low-pass filter with at least 25 dB of attenuation at ω = 2ωc and −1.5 dB at its band edge. How many elements will be required for this filter? If a Chebyshev filter is used with 1.5 dB of ripple in its passband, find the number of circuit elements. 9.6. Design a low-pass Butterworth filter with a cutoff frequency of 150 MHz and an insertion loss of 50 dB at 400 MHz. It is to be used between a 75- load and a generator with its internal resistance as 75 . 9.7. Design a low-pass Chebyshev filter that exhibits ripples no more than 2 dB in its passband. The filter must pass all frequencies up to 50 MHz and attenuate the signal at 100 MHz by at least 15 dB. The load and the source resistance are of 50 each. 9.8. Reconsider Problem 9.7 to design a low-pass filter that exhibits the Chebyshev response with a 3-dB ripple in its passband. The filter must have 50 at both its input and output ports.

390

FILTER DESIGN

9.9. Design a high-pass Chebyshev filter with a passband ripple magnitude below 3 dB. It must pass all frequencies over 50 MHz. A minimum of 15 dB of insertion loss is required at 25 MHz. Assume that the load and source resistances are of 75 each. 9.10. Design a bandpass Chebyshev filter that exhibits no more than 2 dB of ripple in its passband. It must pass the signals that are in the frequency band 50 to 80 MHz with zero insertion loss. Assume that the load and source resistances are of 50 each. 9.11. It is desired to design a maximally flat low-pass filter with at least 40 dB of attenuation at f = 1.1 GHz and −3 dB at f = 1 GHz. How many elements will be required for this filter? If a Chebyshev filter is used with 3 dB of ripple in its passband, find the number of circuit elements. 9.12. Design a Butterworth filter with a cutoff frequency of 960 MHz and an insertion loss of 15 dB at 1.5 GHz. It is to be used between a 50- load and a generator with its internal resistance as 50 . 9.13. Design a low-pass Chebyshev filter that exhibits ripples of no more than 1 dB in its passband. The filter must pass all frequencies up to 960 MHz and attenuate the signal at 1.5 GHz by at least 15 dB. The load and the source resistance are of 50 each. 9.14. Reconsider Problem 9.13 to design a low-pass filter that exhibits the Chebyshev response with 2 dB of ripple in its passband. The filter must have 50 at both its input and output ports. 9.15. Design a high-pass Chebyshev filter with a passband ripple magnitude below 0.5 dB. It must pass all frequencies over 500 MHz. A minimum of 20 dB of insertion loss is required at 200 MHz. Assume that the load and source resistances are of 50 each. 9.16. Design a bandpass Chebyshev filter that exhibits no more than 0.1 dB of ripple in its passband. It must pass the signals that are in the frequency band 500 to 800 MHz with almost zero insertion loss and provide at least 15 dB of attenuation for signals below 300 MHz. Assume that the load and source resistances are of 75 each. 9.17. Design a high-pass filter that exhibits no more than 2.5 dB of ripple in its passband. It must pass all frequencies over 100 MHz. A minimum of 40 dB of insertion loss is required at 25 MHz. Assume that the load and source resistances are of 50 each. 9.18. Design a maximally flat bandstop filter with n = 5. It must stop the signals in the frequency range 50 to 80 MHz and pass the rest of the frequencies. Assume that the load and source resistances are of 50 each.

PROBLEMS

391

9.19. Design a fifth-order low-pass Butterworth filter with a cutoff frequency of 5 GHz. It is to be used between a 75- load and a generator with an internal resistance of 75 . 9.20. Design a third-order low-pass Chebyshev filter that exhibits ripples of no more than 1 dB in its passband. The filter must pass all frequencies up to 1 GHz. The load and the source resistance are of 50 each. 9.21. Design a fifth-order high-pass Chebyshev filter with passband ripple magnitude below 3 dB. It must pass all frequencies over 5 GHz. Assume that the load and source resistances are of 75 each.

10 SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

A signal-flow graph is a graphical means of portraying the relationship among the variables of a set of linear algebraic equations. S. J. Mason originally introduced it to represent the cause and effect of linear systems. Associated terms are defined in this chapter along with a procedure to draw a signal-flow graph for a given set of algebraic equations. Further, signal-flow graphs of microwave networks are obtained in terms of their S-parameters and associated reflection coefficients. The manipulation of signal-flow graphs is summarized to find the desired transfer functions. Finally, the relations for transducer power gain, available power gain, and operating power gain are formulated in this chapter. Consider a linear network that has N input and output ports. It is described by a set of linear algebraic equations: Vi =

N

Zij Ij

i = 1, 2, . . . , N

j =1

This says that the effect Vi at the ith port is a sum of gain times causes at its N ports. Hence, Vi represents the dependent variable (effect), and Ij are the independent variables (cause). Nodes or junction points of the signal-flow graph represent these variables. The nodes are connected by line segments called branches with an arrow on each directed toward the dependent node. The coefficient Zij is the gain of a branch that connects the ith dependent node with the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

392

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

393

j th independent node. A signal can be transmitted through a branch only in the direction of the arrow. The basic properties of a signal-flow graph can be summarized as follows: ž ž ž

ž ž ž

A signal-flow graph can be used only when the system is linear. A set of algebraic equations must be in the form of effects as functions of causes before its signal-flow graph can be drawn. A node is used to represent each variable. Normally, these are arranged from left to right, following a succession of inputs (causes) and outputs (effects) of the network. Nodes are connected together by branches with an arrow directed toward the dependent node. Signals travel along the branches only in the direction of the arrows. Signal Ik traveling along a branch that connects nodes Vi and Ik is multiplied by the branch gain Zik . The dependent node (effect) Vi is equal to the sum of the branch gain times the corresponding independent nodes (causes).

Example 10.1 The output b1 of a system is caused by two inputs a1 and a2 as represented by the equation b1 = S11 a1 + S12 a2 Find its signal-flow graph. SOLUTION There are two independent variables and one dependent variable in the equation. Locate these three nodes and connect nodes a1 and a2 with b1 , as shown in Figure 10.1. The arrows on two branches are directed toward effect b1 . Coefficients of input (cause) are shown as the gain of that branch. Example 10.2 Output b2 of a system is caused by two inputs a1 and a2 as represented by the equation b2 = S21 a1 + S22 a2 Find its signal-flow graph. a1

S11 a2

b1 S12

Figure 10.1

Signal-flow graph representation of Example 10.1.

394

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS a1

S21

a2 b2

Figure 10.2

S22

Signal-flow graph representation of Example 10.2.

SOLUTION There are again two independent variables and one dependent variable in the equation. Locate these three nodes and connect node a1 with b2 and a2 with b2 , as shown in Figure 10.2. The arrows on two branches are directed toward effect b2 . Coefficients of input (cause) are shown as the gain of that branch. Example 10.3 The input–output characteristics of a two-port network are given by the set of linear algebraic equations b1 = S11 a1 + S12 a2 b2 = S21 a1 + S22 a2 Find its signal-flow graph. SOLUTION There are two independent variables a1 and a2 , and two dependent variables b1 and b2 in this set of equations. Locate the four nodes and then connect nodes a1 and a2 with b1 . Similarly, connect a1 and a2 with b2 , as shown in Figure 10.3. Arrows on the branches are directed toward effects b1 and b2 . The coefficients of each input (cause) are shown as the gain of that branch. Example 10.4 The following set of linear algebraic equations represents the input–output relations of a multiport network. Find the corresponding signal-flow graph. S21 a1

b2 S11 S22 a2

b1 S12

Figure 10.3

Signal-flow graph representation of Example 10.3.

395

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

SOLUTION X1 = R1 +

1 X2 2+s

X2 = −4X1 + R2 − 7Y1 −

1 X2 s+4

s X2 +3 Y2 = 10X1 − sY1 Y1 =

s2

In the first equation, R1 and X2 represent the causes and X1 is the effect. Hence, the signal-flow graph representing this equation can be drawn as illustrated in Figure 10.4. Now consider the second equation. X1 is the independent variable in it. Further, X2 appears as cause as well as effect. This means that a branch must start and finish at the X2 node. Hence, when this equation is combined with the first, the signal-flow graph will look as illustrated in Figure 10.5. Next, we add to it the signal-flow graph of the third equation. It has Y1 as the effect and X2 as the cause. It is depicted in Figure 10.6. Finally, the last equation has Y2 as a dependent variable, and X1 and Y1 are two independent variables. A complete signal-flow graph representation is obtained after superimposing it as shown in Figure 10.7.

1 2+s

1 R1

X2 X1

Figure 10.4 Signal-flow graph representation of the first equation of Example 10.4.

−1 s+4 1 1 2+s

1 R1

R2

Y1

X1

X2

−7

−4

Figure 10.5 Signal-flow graph representation of the first two equations of Example 10.4.

396

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS −1 s+4

R2 1

1 2+s

1 R1

X2

X1

−4

Figure 10.6

Y1 −7

s s2 + 3

Signal-flow graph representation of the first three equations of Example 10.4.

−1 s+4 1 1 2+s

1 R1

R2

X1

X2

−7

−4

s s2 + 3

Y1

−s

10 Y2

Figure 10.7 Complete signal-flow graph representation of Example 10.4.

10.1 DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS Before we proceed with manipulation of signal-flow graphs, it will be useful to define a few remaining terms. Input and Output Nodes A node that has only outgoing branches is defined as an input node or source. Similarly, an output node or sink has only incoming branches. For example, R1 , R2 , and Y1 are the input nodes in the signal-flow graph shown in Figure 10.8. This corresponds to the first two equations of Example 10.4. There is no output node (exclude the dotted branches) in it because X1 and X2 have both outgoing as well as incoming branches. Nodes X1 and X2 in Figure 10.8 can be made the output nodes by adding an outgoing branch of

397

DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS

−1 s+4

R2

1 1 2+s

1 R1

Y1

X2

X1

−7 1

1 −4

X1

X2

Figure 10.8 Signal-flow graph with R1 , R2 , and Y1 as input nodes.

unity gain to each node. This is illustrated in Figure 10.8 by dotted branches. It is equivalent to adding X1 = X1 and X2 = X2 in the original set of equations. Thus, any nonoutput node can be made an output node in this way. However, this procedure cannot be used to convert these nodes to input nodes because that changes the equations. If an incoming branch of unity gain is added to node X1 , the corresponding equation is modified as follows: X1 = X1 + R1 +

1 X2 2+s

However, X1 can be made an input node by rearranging it as follows. The corresponding signal-flow graph is illustrated in Figure 10.9. It may be noted that R1 is now an output node: R1 = X1 −

1 X2 2+s

−1 s+4

R2 1

−

R1

1 2+s X2

1

Y1 −7

−4 X1

Figure 10.9 Signal-flow graph with R1 as an output, and X1 , R2 , and Y1 as the input nodes.

398

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Path A continuous succession of branches traversed in the same direction is called the path. It is known as a forward path if it starts at an input node and ends at an output node without hitting a node more than once. The product of branch gains along a path is defined as the path gain. For example, there are two forward paths between nodes X1 and R1 in Figure 10.9. One of these forward paths is just one branch connecting the two nodes with a path gain of 1. The other forward path is X1 to X2 to R1 . Its path gain is 4/(2 + s). Loop A loop is a path that originates and ends at the same node without encountering other nodes more than once along its traverse. When a branch originates and terminates at the same node, it is called a self-loop. The path gain of a loop is defined as the loop gain. Once the signal-flow graph is drawn, the ratio of an output to an input node (while other inputs, if there are more than one, are assumed to be zero) can be obtained by using rules of reduction. Alternatively, Mason’s rule may be used. However, the latter rule is prone to error if the signal-flow graph is too complex. The reduction rules are generally recommended for such cases and are given as follows: Rule 1 When there is only one incoming and one outgoing branch at a node (i.e., two branches are connected in series), it can be replaced by a direct branch with branch gain equal to the product of the two. This is illustrated in Figure 10.10. Rule 2 Two or more parallel paths connecting two nodes can be merged into a single path with a gain that is equal to the sum of the original path gains, as depicted in Figure 10.11. Rule 3 A self-loop of gain G at a node can be eliminated by multiplying its input branches by 1/(1 − G). This is shown graphically in Figure 10.12. Rule 4 A node that has one output and two or more input branches can be split in such a way that each node has just one input and one output branch, as shown in Figure 10.13. 5s

10s

2

X1

X1

X2

R1

X2

Figure 10.10 Graphical illustration of rule 1.

5s 5s + 3 X1

X2

X1

3

Figure 10.11 Graphical illustration of rule 2.

X2

399

DEFINITIONS AND MANIPULATION OF SIGNAL-FLOW GRAPHS

2s 4

X2

X1

4 1 − 2s

X1

X2

Figure 10.12 Graphical illustration of rule 3.

X2

C2

X2

X′4 C4

C2 C4

X1 X4

C1

C1

X1

X5

X′′4

X5 C4

C3

C4

C3

X3

X′′′ 4

X3

Figure 10.13 Graphical illustration of rule 4.

X2

X4 C1

C2

X′′4

C4

C1

C2

X1

X′4

C4 X5

C3

X1

C1

X5

C3

C1 X3

X2

X3 X′′′ 4

Figure 10.14 Graphical illustration of rule 5.

Rule 5 This is similar to rule 4. A node that has one input and two or more output branches can be split in such a way that each node has just one input and one output branch. This is shown in Figure 10.14. Mason’s Gain Rule Ratio T of the effect (output) to that of the cause (input) can be found using Mason’s rule as follows: T (s) =

P1 1 + P2 2 + P3 3 + · · ·

where, Pi is the gain of the ith forward path, =1− L(1) + L(2) − L(3) + · · · 1 = 1 − L(1)(1) + L(2)(1) − L(3)(1) + · · ·

(10.1.1)

(10.1.2) (10.1.3)

400

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

2 = 1 − 3 = 1 −

L(1)(2) +

L(2)(2) −

L(3)(2) + · · ·

(10.1.4)

L(1)(3) + · · ·

(10.1.5)

.. . L(1) stands for the sum of all first-order loop gains, L(2) is the sum of all second-order loop gains, and so on. L(1)(1) denotes the sum of those first-order loop gains that do not touch path P1 at any node, L(2)(1) denotes the sum of those second-order loop gains that do not touch the path of P1 at any point, L(1)(2) denotes the sum of those first-order loops that do not touch path P2 at any point, and so on. First-order loop gain was defined earlier. Second-order loop gain is the product of two first-order loops that do not touch at any point. Similarly, third-order loop gain is the product of three first-order loops that do not touch at any point. Example 10.5 A signal-flow graph of a two-port network is given in Figure 10.15. Using Mason’s rule, find its transfer function Y/R. SOLUTION There are three forward paths from node R to node Y. Corresponding path gains are found as follows: s 3s 1 ×1 ×3×1= s +1 s +2 (s + 1)(s + 2) P2 = 1 × 6 × 1 = 6 P1 = 1 × 1 ×

and P3 = 1 × 1 ×

4 1 × (−4) × 1 = − s +1 s+1

Next, it has two loops. The loop gains are L1 = −

3 s+1

L2 = −

5s s+2

and

6 −4 1/(s + 1)

s/(s + 2)

R

Y 1

1

1 −3

Figure 10.15

3 −5

Signal-flow graph of Example 10.5.

1

SIGNAL-FLOW GRAPH REPRESENTATION OF A VOLTAGE SOURCE

401

Using Mason’s rule, we find that Y P1 + P2 (1 − L1 − L2 + L1 L2 ) + P3 (1 − L2 ) = R 1 − L1 − L2 + L1 L2

10.2 SIGNAL-FLOW GRAPH REPRESENTATION OF A VOLTAGE SOURCE Consider an ideal voltage source ES 0◦ in series with source impedance ZS , as shown in Figure 10.16. It is a single-port network with terminal voltage and current VS and IS , respectively. It is to be noted that the direction of current flow is assumed as entering the port, consistent with that of the two-port networks considered earlier. Further, the incident and reflected waves at this port are assumed to be aS and bS , respectively. The characteristic impedance at the port is assumed to be Z0 . Using the usual circuit analysis procedure, the total terminal voltage VS can be found as follows: VSin − VSref Z0 (10.2.1) where superscripts “in” and “ref” on VS and IS are used to indicate the corresponding incident and reflected quantities. This equation can be rearranged as follows: ZS ZS 1+ VSref = ES − 1 − VSin Z0 Z0 or Z0 Z0 − ZS in VSref = ES − V Z0 + ZS Z0 + ZS S VS = VSin + VSref = ES + ZS IS = ES + ZS (ISin + ISref ) = ES + ZS

IS

ZS

ES ∠0°

~

+

aS

VS

− bS

Figure 10.16 Incident and reflected waves at the output port of a voltage source.

402

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS 1

bS

1 bS

bG ΓS

aS aS

Figure 10.17

Dividing it by

√

1

Signal-flow graph representation for a voltage source.

2Z0 , we find that V ref Z0 Z0 − ZS VSin ES − √S = √ √ Z0 + ZS 2Z0 Z0 + ZS 2Z0 2Z0

Using (8.6.15) and (8.6.16), this can be written as follows: bS = bG + S aS

(10.2.2)

where V ref bS = √ S 2Z0 √ Z0 ES bG = √ 2(Z0 + ZS )

(10.2.3) (10.2.4)

V in aS = √ S 2Z0

(10.2.5)

and S =

ZS − Z0 ZS + Z0

(10.2.6)

From (10.2.2), the signal flow graph for a voltage source can be drawn as shown in Figure 10.17. 10.3 SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE Consider load impedance ZL as shown in Figure 10.18. It is a single-port device with port voltage and current VL and IL , respectively. The incident and reflected waves at the port are assumed to be aL and bL , respectively. Further, characteristic impedance at the port is Z0 . Following the usual circuit analysis rules, we find that VL = VLin + VLref = ZL IL = ZL (ILin + ILref ) = ZL

VLin − VLref Z0

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 403 IL aL

+ ZL

VL −

bL

Figure 10.18 Passive one-port circuit.

or 1+

ZL Z0

VLref =

ZL − 1 VLin Z0

(10.3.1)

or VLref = After dividing by

√

ZL − Z0 in V = L VLin ZL + Z0 L

(10.3.2)

2Z0 , we use (8.6.15) and (8.6.16) to find that bL = L aL

(10.3.3)

where V ref bL = √ L 2Z0

(10.3.4)

V in aL = √ L 2Z0

(10.3.5)

and L =

ZL − Z0 ZL + Z0

(10.3.6)

A signal-flow graph can be drawn on the basis of (10.3.3), as shown in Figure 10.19.

1 aL

aL ΓL

bL

bL 1

Figure 10.19 Signal-flow graph of a one-port passive device.

404

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Γin

Two-port network

ZL

Figure 10.20 Two-port network with termination.

Example 10.6 Impedance ZL terminates port 2 (the output) of a two-port network as shown in Figure 10.20. Draw the signal-flow graph and determine the reflection coefficient at its input port using Mason’s rule. SOLUTION As shown in Figure 10.21, combining the signal-flow graphs of a passive load and that of a two-port network obtained in Example 10.3, we can get the representation for this network. Its input reflection coefficient is given by the ratio of b1 to a1 . For Mason’s rule we find that there are two forward paths, a1 → b1 and a1 → b2 → aL → bL → a2 → b1 . There is one loop, b2 → aL → bL → a2 → b2 , which does not touch the former path. The corresponding path and loop gains are P1 = S11 P2 = S21 × 1 × L × 1 × S12 = L S21 S12 and L1 = 1 × L × 1 × S22 = L S22 Hence, b1 P1 (1 − L1 ) + P2 S11 (1 − L S22 ) + L S21 S12 = = a1 1 − L1 1 − L S22 L S21 S12 = S11 + 1 − L S22

in =

a1

Two-port network S21 b2

S11

Load 1

aL

S22

b1

S12

a2

ΓL

1

bL

Figure 10.21 Signal-flow graph representation of the network shown in Figure 10.20.

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 405 ZS Γout VS ∠0° ~

Two-port network

ZL

Figure 10.22 Two-port network with the source and termination.

Example 10.7 A voltage source is connected at the input port of a two-port network and the load impedance ZL terminates its output, as shown in Figure 10.22. Draw its signal-flow graph and find the output reflection coefficient out . SOLUTION A signal-flow graph of this circuit can be drawn by combining the results of Example 10.6 with those of a voltage source representation obtained in Section 10.2. This is illustrated in Figure 10.23. The output reflection coefficient is defined as the ratio of b2 to a2 with load disconnected, and the source impedance ZS terminates port 1 (the input). Hence, there are two forward paths from a2 to b2 , a2 → b2 , and a2 → b1 → aS → bS → a1 → b2 . There is only one loop (because the load is disconnected), b1 → aS → bS → a1 → b1 . The path and loop gains are found as follows: P1 = S22 P2 = S12 × 1 × S × 1 × S21 = S S12 S21 and L1 = 1 × S × 1 × S11 = S S11 Therefore, from Mason’s rule, we have b2 P1 (1 − L1 ) + P2 S22 (1 − S11 S ) + S21 S12 S = = a2 1 − L1 1 − S11 S S21 S12 S = S22 + 1 − S11 S

out =

Voltage source

Two-port network bS

1

a1

1

S21

Load b2

aL

1

bG ΓS

S11

aS

1

ΓL

S22

b1

S12

a2

1

bL

Figure 10.23 Signal-flow graph representation of the network shown in Figure 10.22.

406

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS Voltage source 1

bS

Load aL

1

bG ΓS

aS

ΓL

1

bL

Figure 10.24 Signal-flow graph for Example 10.8.

Example 10.8 The signal-flow graph shown in Figure 10.24 represents a voltage source that is terminated by a passive load. Analyze the power transfer characteristics of this circuit and establish the conditions for maximum power transfer. SOLUTION Using the defining equations of ai and bi , we can develop the following relations: power output of the source = |bS |2 power reflected back into the source = |aS |2 Hence, power delivered by the source, Pd , is Pd = |bS |2 − |aS |2 Similarly, we can write the following relations for power at the load: power incident on the load = |aL |2 power reflected from the load = |bL |2 = |L |2 |aL |2 where L is the load reflection coefficient. Hence, power absorbed by the load, PL , is given by PL = |aL |2 − |bL |2 = |aL |2 (1 − |L |2 ) = |bS |2 (1 − |L |2 ) Since bS = bG + S aS = bG + S bL = bG + S L aL = bG + S L bS we find that bS =

bG 1 − S L

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 407

and

1 1 aS = (bS − bG ) = S S

bG − bG 1 − S L

=

L bG 1 − S L

Hence, 2 L bG 2 b G − Pd = |bS | − |aS | = 1 − 1 − S L S L 2 bG (1 − |L |2 ) = |bS |2 (1 − |L |2 ) = PL = 1 − S L 2

2

This simply says that the power delivered by the source is equal to the power absorbed by the load. We can use this equation to establish the load condition under which the source delivers maximum power. For that, we consider ways to maximize. 2 bG (1 − |L |2 ) PL = 1− S

L

For PL to be a maximum, its denominator 1 − S L is minimized. Let us analyze this term more carefully using the graphical method. It says that there is a phasor S L at a distance of unity from the origin that rotates with a change in L (S is assumed constant), as illustrated in Figure 10.25. Hence, the denominator has an extreme value whenever S L is a pure real number. If this number is positive real, the denominator is minimum. On the other hand, the denominator is maximized when S L is a negative real number. For S L to be a positive real number, the load reflection coefficient L must be the complex conjugate of the source reflection coefficient S . In other words, L = S∗ ⇒

ZL − Z0 = ZL + Z0

ZS − Z0 ZS + Z0

∗

1 ΓSΓL

1 − ΓSΓL

Figure 10.25 Graphical representation of (1 − S L ).

408

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

For a real Z0 it reduces to

ZL = ZS∗

The value of this maximum power is PL =

|bG |2 = Pavs (1 − |S |2 )

where Pavs is the maximum power available from the source. It is a well-known maximum power transfer theorem of electrical circuit theory. Example 10.9 Draw the signal-flow graph of the three-port network shown in Figure 10.26. Find the transfer characteristics b3 /bS using the graph. SOLUTION For a three-port network, the scattering equations are given as follows: b1 = S11 a1 + S12 a2 + S13 a3 b2 = S21 a1 + S22 a2 + S23 a3 b3 = S31 a1 + S32 a2 + S33 a3 For the voltage source connected at port 1, we have bS = bG + S aS Since the wave coming out from the source is incident on port 1, bS is equal to a1 . Therefore, we can write a1 = bG + S b1 For the load ZL connected at port 2, bL = L aL

ZD ZS

VS ∠0° ~

Port 1

Port 3

Three-port network

Port 2

Figure 10.26 Three-port network of Example 10.9.

ZL

SIGNAL-FLOW GRAPH REPRESENTATION OF A PASSIVE SINGLE-PORT DEVICE 409

Again, the wave emerging from port 2 of the network is incident on the load. Similarly, the wave reflected back from the load is incident on port 2 of the network. Hence, b2 = aL and bL = a2 Therefore, the relation above at the load ZL can be modified as follows: a2 = L b2 Following a similar procedure for the load ZD connected at port 3, we find that bD = D aD ⇒ a3 = D b3 Using these equations, a signal-flow graph can be drawn for this circuit as shown in Figure 10.27. For b3 /bs , there are two forward paths and eight loops in this flow graph. Path gains Pi and loop gains Li are found as follows: P1 = S31 P2 = S21 S32 L ΓD

b3

a3 S33

S31

S23 S13

1

bS

S32

a1

b2 1

aL

bG 1 ΓS

aS

Figure 10.27

S21

S11

1

ΓL

S22

b1

S12

a2

1

bL

Signal-flow graph of the network shown in Figure 10.26.

410

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

L1 = S11 S L2 = S22 L L3 = S33 D L4 = S21 S12 L S L5 = S31 S23 S12 D L S L6 = S13 S21 S32 D L S L7 = S23 S32 D L and L8 = S13 S31 D S Therefore, various terms of Mason’s rule can be evaluated as follows: 1 = 1 − S22 L 2 = 1 L(1) = L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8 L(2) = L1 L2 + L1 L3 + L2 L3 + L3 L4 + L1 L7 + L2 L8 L(3) = L1 L2 L3 Using Mason’s rule we can find that b3 P1 1 + P2 2 = bS 1 − L(1) + L(2) − L(3) Note that this signal-flow graph is too complex, and therefore one can easily miss a few loops. In cases like this, it is prudent to simplify the signal-flow graph using the five rules mentioned earlier.

10.4 POWER GAIN EQUATIONS We have seen in the preceding section that maximum power is transferred when load reflection coefficient L is the conjugate of the source reflection coefficient S . It can be generalized for any port of the network. In the case of an amplifier, maximum power will be applied to its input port if its input reflection coefficient in is the complex conjugate of the source reflection coefficient S . If this is not the case, the input will be less than the maximum power available from the source. Similarly, maximum amplified power will be transferred to the load only if load reflection coefficient L is the complex conjugate of output reflection coefficient out . Part of the power available at the output of the amplifier will be

411

POWER GAIN EQUATIONS

reflected back if there is a mismatch. Therefore, the power gain of an amplifier can be defined at least three different ways as follows: PL power delivered to the load = Pavs power available from the source

transducer power gain, GT = operating power gain, GP =

PL power delivered to the load = Pin power input to the network

available power gain, GA =

PAVN power available from the network = Pavs power available from the source

and

where Pavs is the power available from the source, Pin the power input to the network, PAVN the power available from the network, and PL the power delivered to the load. Therefore, the transducer power gain will be equal to that of operating power gain if the input reflection coefficient in is the complex conjugate of the source reflection coefficient S . That is, Pavs = Pin |in =S∗ Similarly, the available power gain will be equal to the transducer power gain if the following condition holds at the output port: ∗ PAVN = PL |L =out

In this section we formulate these power gain relations in terms of scattering parameters and reflection coefficients of the circuit. To facilitate the formulation, we reproduce in Figure 10.28 the signal-flow graph obtained in Example 10.7.

Amplifier circuit (two-port)

Voltage source 1

bS

a1

1

S21

Load 1

b2

aL

bG S11

ΓS

aS

Pavs

1

Pin

ΓL

S22

b1

S12

a2

PAVN

1

bL

PL

Figure 10.28 Signal-flow graph for an amplifier circuit with voltage source and terminating load.

412

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Transducer Power Gain To develop the relation for transducer power gain, we first determine the power delivered to load PL and the power available from the source Pavs : PL = |aL |2 − |bL |2 = |aL |2 (1 − |L |2 ) = |b2 |2 (1 − |L |2 )

(10.4.1)

We derived an expression for Pavs in Example 10.7. That is rewritten here: Pavs = |bS |2 − |aS |2 =

|bG |2 1 − |S |2

(10.4.2)

Therefore, the transducer power gain is GT =

|b2 |2 (1 − |L |2 )(1 − |S |2 ) |bG |2

(10.4.3)

We now need to find an expression for b2 /bG . Mason’s rule can be used to formulate it. There is only one forward path in this case. However, there are three loops. The path gain P1 and loop gains Li are P1 = S21 L1 = S11 S L2 = S22 L and L3 = S12 S21 S L Therefore, P1 b2 = bG 1 − (L1 + L2 + L3 ) + L1 L2 S21 = 1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L

(10.4.4)

Substituting (10.4.4) into (10.4.3), we get GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L |2

It can be simplified as follows: GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S11 S )(1 − S22 L ) − S21 S12 L S |2

(10.4.5)

413

POWER GAIN EQUATIONS

or GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S22 L )[1 − S11 S − S21 S12 L S /(1 − S22 L )]|2

GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) |(1 − S22 L )(1 − in S )|2

or

=

1 − |S |2 1 − |L |2 2 |S | 21 |1 − in S |2 |1 − S22 L |2

where in = S11 +

S21 S12 L 1 − S22 L

(10.4.6)

(10.4.7)

This is the input reflection coefficient formulated in Example 10.6. Alternatively, (10.4.5) can be written as GT =

|S21 |2 (1 − |L |2 )(1 − |S |2 ) 1 − |S |2 1 − |L |2 = |S21 |2 2 2 |(1 − S11 S )(1 − out L )| |1 − S11 S | |1 − out L |2 (10.4.8)

where out = S22 +

S21 S12 S 1 − S11 S

(10.4.9)

This is the output reflection coefficient formulated in Example 10.7. Special Case Consider the case when the two-port network is unilateral. This means that there is an output signal at port 2 when a source is connected at port 1. However, a source (cause) connected at port 2 does not show its response (effect) at port 1. In terms of scattering parameters of the two-port, S12 is equal to zero in this case, and therefore, the input and output reflection coefficients in (10.4.7) and (10.4.9) simplify as follows: in = S11 and

out = S22

The transducer power gain GT as given by (10.4.6) or (10.4.8) simplifies to GT |S12 =0 = GT U =

1 − |S |2 1 − |L |2 2 |S | = GS G0 GL 21 |1 − S11 S |2 |1 − S22 L |2

(10.4.10)

where GT U may be called the unilateral transducer power gain, while GS =

1 − |S |2 |1 − S11 S |2

G0 = |S21 |2

(10.4.11) (10.4.12)

414

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

and GL =

1 − |L |2 |1 − S22 L |2

(10.4.13)

Equation (10.4.10) indicates that maximum GT U is obtained when the input and output ports are conjugate matched. Hence, GT Umax = GSmax G0 GLmax

(10.4.14)

GSmax = GS |S =S11∗

(10.4.15)

GLmax = GL |L =S22∗

(10.4.16)

where

and

Operating Power Gain To find the operating power gain, we require the power delivered to the load and power input to the circuit. We evaluated the power delivered to the load in (10.4.1). The power input to the circuit is found as follows: Pin = |a1 |2 − |b1 |2 = |a1 |2 (1 − |in |2 ) Hence, GP =

PL |b2 |2 (1 − |L |2 ) = Pin |a1 |2 (1 − |in |2 )

(10.4.17)

Now we evaluate b2 /a1 using Mason’s rule. With the source disconnected, we find that there is only one forward path and one loop with gains S21 and S22 L , respectively. Hence, b2 S21 = (10.4.18) a1 1 − S22 L Substituting (10.4.18) into (10.4.17), we find that GP =

1 1 − |L |2 |S21 |2 2 1 − |in | |1 − S22 L |2

(10.4.19)

Available Power Gain Expressions for maximum powers available from the network and the source are needed to determine the available power gain. Maximum power available from the network PAVN is delivered to the load if the load reflection coefficient is a complex conjugate of the output. Using (10.4.1), we find that 2 2 2 2 ∗ = |b | (1 − | | ) = |b | (1 − | PAVN = PL |L =out 2 L 2 out | )

(10.4.20)

415

POWER GAIN EQUATIONS

The power available from the source is given by (10.4.2). Hence, GA =

PAVN |b2 |2 = (1 − |out |2 )(1 − |S |2 ) Pavs |bG |2

(10.4.21)

Next, using Mason’s rule in the signal-flow graph of Figure 10.28, we find that S21 b2 = bG 1 − (S11 S + S22 L + S21 L S12 S ) + S11 S S22 L S21 = (1 − S11 S )(1 − out L ) and

S21 b2 = bG L =out (1 − S11 S )(1 − |out |2 ) ∗

(10.4.22)

Now, substituting (10.4.22) into (10.4.21), we get GA =

1 − |S |2 1 |S21 |2 2 |1 − S11 S | 1 − |out |2

(10.4.23)

Example 10.10 The scattering parameters of a microwave amplifier are found at 800 MHz as follows (Z0 = 50 ): S11 = 0.45 150◦ , S12 = 0.01 − 10◦ , S21 = 2.05 10◦ , and S22 = 0.4 − 150◦ . The source and load impedances are 20 and 30 , respectively. Determine the transducer power gain, operating power gain, and available power gain. SOLUTION First we determine the source reflection coefficient S and the load reflection coefficient L as follows: S =

ZS − Z0 20 − 50 30 = =− = −0.429 ZS + Z0 20 + 50 70

L =

ZL − Z0 30 − 50 20 = =− = −0.25 ZL + Z0 30 + 50 80

and

Now the input reflection coefficient in and the output reflection coefficient out can be calculated from (10.4.7) and (10.4.9), respectively: in = S11 + or

S21 S12 L (0.01 − 10◦ )(2.05 10◦ )(−0.25) ◦ = 0.45 150 + 1 − S22 L 1 − (0.4 − 150◦ )(−0.25) ◦

in = −0.3953 + j 0.2253 = 0.455 150.32

416

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

and S21 S12 S = 0.4 1 − S11 S

out = S22 +

◦

− 150 +

(0.01 − 10◦ )(2.05 10◦ )(−0.429) 1 − (0.45 150◦ )(−0.429)

or out = −0.3568 − j 0.1988 = 0.4084

◦

− 150.87

The transducer power gain GT can now be calculated from (10.4.6) or (10.4.8) as follows: |S21 |2 (1 − |L |2 )(1 − |S |2 ) |S21 |2 (1 − |L |2 )(1 − |S |2 ) = |(1 − S22 L )(1 − in S )|2 |(1 − S11 S )(1 − out L )|2

GT = Hence,

1 − (0.429)2 1 − (0.252 ) 2 (2.05) |1 − (0.45 150◦ )(−0.429)|2 |1 − (0.4084 − 150.87◦ )(−0.25)|2

GT =

= 5.4872 Similarly, from (10.4.19), GP =

1 1 − |L |2 2 |S | 21 1 − |in |2 |1 − S22 L |2

Therefore, GP =

1 1 − (0.252 ) 2 ) = 5.9374 (2.05 1 − (0.4552 ) |1 − (0.4 − 150◦ )(−0.25)|2

The available power gain is calculated from (10.4.23) as follows: GA =

1 − |S |2 1 |S21 |2 2 |1 − S11 S | 1 − |out |2

Hence, GA =

1 − (0.429)2 1 (2.052 ) = 5.8552 |1 − (0.45 150◦ )(−0.429)|2 1 − (0.4084)2

The three power gains can be expressed in decibels as follows: GT (dB) = 10 log(5.4872) ≈ 7.4 dB GP (dB) = 10 log(5.9374) ≈ 7.7 dB and GA (dB) = 10 log(5.8552) ≈ 7.7 dB

417

SUGGESTED READING

Example 10.11 The scattering parameters of a microwave amplifier are found at 2 GHz as follows (Z0 = 50 ): S11 = 0.97 − 43◦ , S12 = 0.0, S21 = 3.39 140◦ , and S22 = 0.63 − 32◦ . The source and load reflection coefficients are 0.97 43◦ and 0.63 32◦ , respectively. Determine the transducer power gain, operating power gain, and available power gain. SOLUTION Since S12 is zero, this amplifier is unilateral. Therefore, its input and output reflection coefficients from (10.4.7) and (10.4.9) may be found as follows: ◦

input reflection coefficient in = S11 = 0.97

− 43

output reflection coefficient out = S22 = 0.63

− 32

◦

On comparing the input reflection coefficient with that of the source, we find that one is the complex conjugate of the other. Hence, the source is matched with the input port, and therefore the power available from the source is equal to the power input. Similarly, we find that the load reflection coefficient is the conjugate of the output reflection coefficient. Hence, power delivered to the load will be equal to power available from the network at port 2. As may be verified from (10.4.10), (10.4.19), and (10.4.23), these conditions make all three power gains equal. From (10.4.10), GT U =

1 − |L |2 1 − |S |2 1 − (0.97)2 2 |S | = (3.39)2 21 |1 − S11 S |2 |1 − S22 L |2 |1 − (0.97)2 |2 ×

1 − (0.63)2 = 322.42 |1 − (0.63)2 |2

From (10.4.19), GP =

2 1 1 1 − |L |2 2 2 1 − (0.63) |S | = (3.39) 21 1 − |in |2 |1 − S22 L |2 1 − (0.97)2 |1 − (0.63)2 |2

= 322.42 From (10.4.23), GA =

1 − |S |2 1 − (0.97)2 1 1 |S21 |2 = (3.39)2 2 2 2 |1 − S11 S | 1 − |out | |1 − (0.97) | 1 − (0.63)2

= 322.42

SUGGESTED READING Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Davis, W. A., Microwave Semiconductor Circuit Design. New York: Van Nostrand Reinhold, 1984.

418

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

Fusco, V. F., Microwave Circuits. Englewood Cliffs, NJ: Prentice Hall, 1987. Gonzalez, G., Microwave Transistor Amplifiers. Upper Saddle River, NJ: Prentice Hall, 1997. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Wolff, E. A., and R. Kaul, Microwave Engineering and Systems Applications. New York: Wiley, 1988.

PROBLEMS 10.1. Use Mason’s rule to find Y (s)/R(s) for the signal-flow graph shown in Figure P10.1. 3

1

2

1/S

1

1/S

1 Y(s)

R(s)

−4 −5

Figure P10.1

10.2. Use Mason’s gain rule to find Y5 (s)/Y1 (s) and Y2 (s)/Y1 (s) for the system shown in Figure P10.2. G4

1

Y2

G1

Y3

G2

Y4

G3

Y6

1 Y5

Y1 −H1

−H2

−H3

Figure P10.2

419

PROBLEMS

10.3. Use Mason’s gain rule to find Y2 (s)/R3 (s) for the system shown in Figure P10.3. R2(s) R3(s)

1

1/S

Y3

Y4

6

Y6

1/S

1 Y1

R1(s) −3

−2 Y2(s) −4

Figure P10.3

10.4. Use Mason’s gain rule to find Y (s)/R(s) for the signal-flow graph shown in Figure P10.4. R(s)

1

4 3

2

1/(S + 1)

1 1/(S + 1) 6 7

5

Y(s) 1

Figure P10.4

10.5. Draw the signal-flow graph for the circuit shown in Figure P10.5. Use Mason’s gain rule to determine the power ratios P2 /P1 and P3 /P1 . Port 3 P3 Port 1

[S]

P1

Port 2

Figure P10.5

0 [S] = S21 0

S12 0 S32

0 S23 0

P2

420

SIGNAL-FLOW GRAPHS AND THEIR APPLICATIONS

in is the reflection coefficient at port 1, and 2 and 3 are the reflection coefficients at ports 2 and 3, respectively. 10.6. A voltage source has an internal impedance of 100 , and an electromotive force of 3 V. Using Z0 = 50 , find the reflection coefficient and scattering variables a and b for the source. 10.7. A 10-mV voltage source has an internal impedance of 100 + j 50 . Assuming that the characteristic impedance Z0 is 50 , find the reflection coefficient and scattering variables a and b for the source. Draw the signal-flow graph. 10.8. A voltage source of 1 0◦ V has an internal resistance of 80 . It is connected to a load of 30 + j 40 . Assuming that the characteristic impedance is 50 , find the scattering variables and draw the signal-flow graph. 10.9. The scattering parameters of a two-port network are S11 = 0.26 − j 0.16, S12 = S21 = 0.42, and S22 = 0.36 − j 0.57. (a) Determine the input reflection coefficient and transducer power loss for ZS = ZL = Z0 . (b) Do the same for ZS = Z0 and ZL = 3Z0 . 10.10. Two two-port networks are connected in cascade. The scattering parameters of the first two-port network are S11 = 0.28, S12 = S21 = 0.56, and S22 = 0.28 − j 0.56. For the second network, S11 = S22 = −j 0.54 and S12 = S21 = 0.84. Determine the overall scattering and transmission parameters. 10.11. Draw the flow graph representation for the two cascaded circuits of Problem 10.10. Using Mason’s rule, determine the following: (a) the input reflection coefficient when the second network is match-terminated and (b) the forward transmission coefficient of the overall network. 10.12. A GaAs MESFET has the following S-parameters measured at 8 GHz with a 50- reference: S11 = 0.26 − 55◦ , S12 = 0.08 80◦ , S21 = 2.14 65◦ , and S22 = 0.82 − 30◦ . For ZL = 40 + j 160 and S = 0.31 − 76.87◦ , determine the operating power gain of the circuit. 10.13. The S-parameters of a certain microwave transistor measured at 3 GHz with a 50- reference are S11 = 0.406 − 100◦ , S12 = 0, S21 = 5.0 50◦ , and S22 = 0.9 − 60◦ . Calculate the maximum unilateral power gain. 10.14. (a) Find the value of the source impedance that results in maximum power delivered to the load in the circuit shown in Figure P10.14. Evaluate the maximum power delivered to the load.

421

PROBLEMS Zs

10 V ∠0°

Z0 = 50 Ω

ZL = 50 + j50 Ω

λ /4

Figure P10.14

(b) Using the value of Zs from part (a), find the Th´evenin equivalent circuit at the load end and evaluate the power delivered to the load.

11 TRANSISTOR AMPLIFIER DESIGN

Amplifiers are among the basic building blocks of an electronic system. Although vacuum-tube devices are still used in high-power microwave circuits, transistors—silicon bipolar junction devices, GaAs MESFETs, heterojunction bipolar transistors (HBTs), and high-electron mobility transistors (HEMTs)—are common in many radio-frequency and microwave designs. The chapter begins with the stability considerations for a two-port network and the formulation of relevant conditions in terms of its scattering parameters. Expressions for input and output stability circles are presented next to facilitate the design of amplifier circuits. Design procedures for various small-signal single-stage amplifiers are discussed for unilateral as well as bilateral transistors. Noise figure considerations in amplifier design are discussed in the following section. An overview of broadband amplifiers is included. Small-signal equivalent circuits and biasing mechanisms for various transistors are also summarized in subsequent sections.

11.1 STABILITY CONSIDERATIONS Consider a two-port network that is terminated by load ZL as shown in Figure 11.1. A voltage source VS with internal impedance ZS is connected at its input port. Reflection coefficients at its input and output ports are in and out , respectively. The source reflection coefficient is S and the load reflection coefficient is L . Expressions for input and output reflection coefficients were formulated in Examples 10.6 and 10.7. For this two-port to be unconditionally stable at a given frequency, the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

422

423

STABILITY CONSIDERATIONS ZS

VS ∠0°

~

ZL

Two-port network

ΓS Zin, Γin

Zout, Γout ΓL

Figure 11.1 Two-port network with voltage source at its input and a load terminating the output port.

following inequalities must hold:

and

|S | < 1

(11.1.1)

|L | < 1 S21 S12 L 0 or 1 − |S11 |2 − |S22 |2 + ||2 > 2|S12 S21 | or

1 − |S11 |2 − |S22 |2 + ||2 >1 2|S12 S21 |

Therefore, k=

1 − |S11 |2 − |S22 |2 + ||2 >1 2|S12 S21 |

(11.1.18)

426

TRANSISTOR AMPLIFIER DESIGN

If the S-parameters of a transistor satisfy conditions (11.1.16) and (11.1.18), it is stable for any passive load and generator impedance. In other words, this transistor is unconditionally stable. On the other hand, it may be conditionally stable (stable for limited values of load or source impedance) if one or both of these conditions are violated. It means that the transistor can provide stable operation for a restricted range of S and L . A simple procedure to find these stable regions is to test inequalities (11.1.3) and (11.1.4) for particular load and source impedances. An alternative graphical approach is to find the circles of instability for load and generator reflection coefficients on a Smith chart. The latter approach is presented below. From the expression of input reflection coefficient (10.4.7), we find that in = S11 +

S21 S12 L ⇒ in (1 − S22 L ) = S11 (1 − S22 L ) + S21 S12 L 1 − S22 L

or in = S11 − L (S11 S22 − S12 S21 − in S22 ) ⇒ L =

S11 − in − in S22

or L =

S11 − in S22 1 S11 S22 − in S22 − S12 S21 + S12 S21 = − in S22 S22 S22 − in S22

L =

1 S22

or

1+

S12 S21 − in S22

=

1 S22

+

S12 S21 1 − in −1 S22

(11.1.19)

As before, 1 − in S22 −1 represents a circle on the complex plane. It is centered at 1 with radius |in S22 −1 |; the reciprocal of this expression is another circle with center at 1 1 1 1 + = −1 −1 2 1 + | S22 | 1 − | S22 | 1 − |−1 S22 |2 and radius 1 2

1 1 − 1 − |−1 S22 | 1 + |−1 S22 |

=

|−1 S22 | 1 − |−1 S22 |2

Since |in | < 1, the region of stability will include all points on the Smith chart outside this circle. From (11.1.19), the center of the load impedance circle, CL , is CL =

1 S22

or 1 CL = S22

+

S12 S21 1 − |−1 S22 |2

S12 S21 ∗ 1+ ||2 − |S22 |2

=

=

1 S22

+

S12 S21 ||2 ||2 − |S22 |2

1 ||2 − |S22 |2 + S12 S21 ∗ S22 ||2 − |S22 |2

STABILITY CONSIDERATIONS

or CL =

427

1 ∗ ( + S12 S21 ) − |S22 |2 1 ∗ S11 S22 − |S22 |2 = S22 ||2 − |S22 |2 S22 ||2 − |S22 |2

Therefore, CL =

∗ ∗ ∗ ∗ S11 − S22 (S22 − S11 ) = ||2 − |S22 |2 |S22 |2 − ||2

(11.1.20)

Its radius, rL , is given by 1 S12 S21 |−1 S22 | S12 S21 = rL = |S22 | 1 − |−1 S22 |2 ||2 − |S22 |2

(11.1.21)

As explained following (11.1.19), this circle represents the locus of points over which the input reflection coefficient in is equal to unity. On one side of this circle, the input reflection coefficient is less than unity (stable region); on its other side it exceeds 1 (unstable region). When load reflection coefficient L is zero (i.e., a matched termination is used), in is equal to S11 . Hence, the center of the Smith chart (reflection coefficient equal to zero) represents a stable point if |S11 | is less than unity. On the other hand, it represents unstable impedance for |S11 | greater than unity. If L = 0 is located outside the stability circle and is found stable, all outside points are stable. Similarly, if L = 0 is inside the stability circle and is found stable, all enclosed points are stable. If L = 0 is unstable, all points on that side of the stability circle are unstable. Similarly, the locus of S can be derived from (11.1.4), with its center CS and its radius rS given as CS = and rS =

∗ ∗ ∗ ∗ S22 − S11 (S11 − S22 ) = 2 2 2 || − |S11 | |S11 | − ||2

1 S12 S21 |−1 S11 | S12 S21 = |S11 | 1 − |−1 S11 |2 ||2 − |S11 |2

(11.1.22)

(11.1.23)

This circle represents the locus of points over which output reflection coefficient out is equal to unity. On one side of this circle, the output reflection coefficient is less than unity (stable region), whereas on its other side it exceeds 1 (unstable region). When the source reflection coefficient S is zero, out is equal to S22 . Hence, the center of the Smith chart (reflection coefficient equal to zero) represents a stable point if |S22 | is less than unity. On the other hand, it represents an unstable impedance point for |S22 | greater than unity. If S = 0 is located outside the stability circle and is found stable, all outside source-impedance points are stable. Similarly, if S = 0 is inside the stability circle and is found stable, all enclosed points are stable. If S = 0 is unstable, all points on that side of the stability circle are unstable.

428

TRANSISTOR AMPLIFIER DESIGN

Example 11.1 The S-parameters of a properly biased transistor are found at ◦ 2 GHz as follows (50- reference impedance): S11 = 0.894 −60.6 , S12 = ◦ ◦ ◦ 0.02 62.4 , S21 = 3.122 123.6 , and S22 = 0.781 −27.6 . Determine its stability and plot the stability circles if the transistor is potentially unstable (see Figure 11.3). SOLUTION From (11.1.16) and (11.1.18), we get || = |S11 S22 − S12 S21 | = 0.6964 and

1 + ||2 − |S11 |2 − |S22 |2 = 0.6071 2|S12 S21 |

k=

Since one of the conditions for stability failed above, this transistor is potentially unstable.

0 12

110

90

100

rS

80

CL

70 CS 60 50

0

rL

10

170

20

160

15

30

0

14

40

0

13

10

5

2

1

0.5

−10

70 0 −1

0.2

0

180

0

50 −1

−16

−20 0

−1

−4

40

−3 0

−5

0

30

−6

0

−70

20

−80

0

−90

−100 −11

−1

−1

Figure 11.3

Input and output stability circles for Example 11.1.

AMPLIFIER DESIGN FOR MAXIMUM GAIN

429

Using (11.1.20) to (11.1.23), we can determine the stability circles as follows. For the output stability circle CL = 1.36 46.7◦ and rL = 0.5. Since |S11 | is 0.894, L = 0 represents a stable load point. Further, this point is located outside the stability circle, and therefore all points outside this circle are stable. For the input stability circle CS = 1.13 68.5◦ and rS = 0.2. Since |S22 | is 0.781, S = 0 represents a stable source impedance point. Further, this point is located outside the stability circle, and therefore all points outside this circle are stable. For the output stability circle, draw a radial line at 46.7◦ . With the radius of the Smith chart as unity, locate the center at 1.36 on this line. It can be done as follows. Measure the radius of the Smith chart using a ruler scale. Supposing that it is d millimeters, the location of the stability circle is then at 1.36d millimeters away on this radial line. Similarly, the radius of the stability circle is 0.5d millimeters. The load impedance must lie outside this circle for the circuit to be stable. Following a similar procedure, the input stability circle is drawn with its radius as 0.2d millimeters and center at 1.13d millimeters on the radial line at 68.5◦ . To have a stable design, the source impedance must lie outside this circle.

11.2 AMPLIFIER DESIGN FOR MAXIMUM GAIN In this section we first consider the design of an amplifier that uses a unilateral transistor (S12 = 0) and has maximum possible gain. A design procedure using a bilateral transistor (S12 = 0) is developed next that requires simultaneous conjugate matching at its two ports. Unilateral Case When S12 is zero, the input reflection coefficient in reduces to S11 and the output reflection coefficient out simplifies to S22 . To obtain maximum gain, the ∗ ∗ and S22 , respectively. source and load reflection coefficients must be equal to S11 Further, the stability conditions for a unilateral transistor simplify to |S11 | < 1 and |S22 | < 1. Example 11.2 The S-parameters of a properly biased BJT are found at 1 GHz ◦ ◦ as follows (with Z0 = 50 ): S11 = 0.60 −155 , S22 = 0.48 −20 , S12 = 0, ◦ and S21 = 6 180 . Determine the maximum gain possible with this transistor and design an RF circuit that can provide this gain. SOLUTION 1. Stability check: k=

1 − |S11 |2 − |S22 |2 + ||2 =∞ 2|S12 S21 |

430

TRANSISTOR AMPLIFIER DESIGN

because S12 = 0 and || = |S11 S22 − S12 S21 | = |S11 S22 | = 0.2909 Since both of the conditions are satisfied, the transistor is unconditionally stable. 2. The maximum possible power gain of the transistor is found as GTU =

2 1 − |s |2 2 1 − |L | |S | 21 |1 − S11 s |2 |1 − S22 L |2

and ∗ 2 ∗ 2 1 − |S11 | 2 1 − |S22 | |S | 21 |1 − |S11 |2 |2 |1 − |S22 |2 |2 2

1 1 = = 73.9257 6 1 − 0.6062 1 − 0.482

GT Umax =

or GTUmax = 10 log10 (73.9257) dB = 18.688 dB 3. For the maximum unilateral power gain, ◦

∗ S = S11 = 0.606 155

◦

∗ and L = S22 = 0.48 20

The component values are determined from the Smith chart as illustrated in Figure 11.4. Corresponding to the reflection coefficient’s magnitude 0.606, the VSWR is 4.08. Hence, point D in Figure 11.4 represents the source impedance ZS . Similarly, point B can be identified as the load impedance ZL . Alternatively, the normalized impedance can be computed from the reflection coefficient. Then, points D and B can be located on the Smith chart. Now, we need to transform the zero reflection coefficient (normalized impedance of 1) to S (point D) at the input and to L (point B) at the output. One way to achieve this is to move first on the unity conductance circle from the center of the chart to point C and then on the constant-resistance circle to reach point D. Thus, it requires a shunt capacitor and then a series inductor to match at the input. Similarly, we can follow the unity-resistance circle from 1 to point A and then the constant-conductance circle to reach point B. Thus, a capacitor is connected in series with the load and then an inductor in shunt to match its output side. Actual values of the components are determined as follows. The normalized susceptance at point C is estimated as j 1.7. Hence, the shunt capacitor on the source side must provide a susceptance of 1.7/50 = 0.034 S.

431

AMPLIFIER DESIGN FOR MAXIMUM GAIN

0

110

90

100

80

70 60

12

50

30

20

160

15

30

0

14

40

0

1

170

10

D B

10

5

2

1

0.5

−10

70 0 −1 −16

0.2

0

180

0

A 0

−1

−4

40

−

0

−3

0 15

−20

C

−5

30

0

−6

0

−70

20

−80

0

−90

−100 −11

−1

−1

Figure 11.4

Smith chart illustrating the design of Example 11.2.

2.307 pF

50 Ω

5.1725 nH 12.434 nH

~

50 Ω

5.411 pF

Figure 11.5 RF circuit designed for Example 11.2.

Since the signal frequency is 1 GHz, this capacitance must be 5.411 pF. Now the change in normalized reactance from point C to point D is determined as j 0.2 − (−j 0.45) = j 0.65. The positive sign indicates that it is an inductor of reactance 0.65 × 50 = 32.5 . The corresponding inductance is found as 5.1725 nH. Similarly, the normalized reactance at point A is estimated as −j 1.38. Hence, the load requires a series capacitor of 2.307 pF. For transforming the susceptance of point A to that of point B, we need an inductance of (−0.16 − 0.48)/50 = −0.0128 S. It is found to be a shunt inductor of 12.434 nH. This circuit is illustrated in Figure 11.5.

432

TRANSISTOR AMPLIFIER DESIGN

Bilateral Case (Simultaneous Conjugate Matching) To obtain the maximum possible gain, a bilateral transistor must be matched at both of its ports simultaneously, as shown in Figure 11.6. When its output port is properly terminated, the input side of the transistor is matched with the source such that in = S∗ . Similarly, the output port of transistor is matched with the load when its input is matched terminated. Mathematically, in = S∗ and

out = L∗

Therefore, S∗ = S11 +

S12 S21 L 1 − S22 L

(11.2.1)

L∗ = S22 +

S12 S21 S 1 − S11 S

(11.2.2)

∗ ∗ S12 S21 ∗ ∗ (1/ L ) − S22

(11.2.3)

and

From (11.2.1), ∗ + S = S11

and from (11.2.2), L∗ =

S22 − (S11 S22 − S12 S21 )S S22 − S = 1 − S11 S 1 − S11 S

(11.2.4)

Substituting (11.2.4) in (11.2.3), we have ∗ + S = S11

∗ ∗ S12 S21 ∗ [(1 − S11 S )/(S22 − S )] − S22

ZS

VS ∠0°

~

Input matching network

Output match- Load ing ZL network

Transistor

ΓS

Γin

Γout

ΓL

Figure 11.6 Bilateral transistor with input- and output-matching networks.

433

AMPLIFIER DESIGN FOR MAXIMUM GAIN

or ∗ + S = S11

∗ ∗ ∗ ∗ S12 S12 S21 (S22 − S ) S21 (S22 − S ) ∗ + = S 11 ∗ ∗ 1 − S11 S − |S22 |2 + S S22 1 − |S22 |2 − (S11 − S22 )S

or ∗ ∗ ∗ ∗ (1 − |S22 |2 )S − (S11 − S22 )S2 = S11 (1 − |S22 |2 ) − S11 (S11 − S22 )S ∗ ∗ + S12 S21 (S22 − S )

or ∗ ∗ (S11 − S22 )S2 + (||2 − |S11 |2 + |S22 |2 − 1)S + (S11 − S22 ∗ ) = 0

This is a quadratic equation in S and its solution can be found as follows: S =

B1 ±

B12 − 4|C1 |2 2C1

= MS

(11.2.5)

where B1 = 1 + |S11 |2 − |S22 |2 − ||2 and

∗ C1 = S11 − S22

Similarly, a quadratic equation for L can be formulated. Solutions to that equation are found to be ML =

B2 ±

B22 − 4|C2 |2 2C2

(11.2.6)

where B2 = 1 + |S22 |2 − |S11 |2 − ||2 and ∗ C2 = S22 − S11

If |B1 /2C1 | > 1 and B1 > 0 in equation (11.2.5), the solution with a minus sign produces |MS | < 1 and that with a positive sign produces |MS | > 1. On other hand, if |B1 /2C1 | > 1 with B1 < 0 in this equation, the solution with a plus sign produces |MS | < 1 and the solution with a minus sign produces |MS | > 1. Similar considerations apply to equation (11.2.6) as well. Observation

|Bi /2Ci | > 1 implies that |k| > 1.

434

TRANSISTOR AMPLIFIER DESIGN

Proof: Bi 2C

i

or

2 2 2 > 1 ⇒ 1 + |S11 | − |S22 | − || > 1 2(S − S ∗ ) 11

22

∗ |1 + |S11 |2 − |S22 |2 − ||2 | > 2|(S11 − S22 )|

Squaring on both sides of this inequality, we get ∗ |1 + |S11 |2 − |S22 |2 − ||2 |2 > 4|(S11 − S22 )|2

and ∗ ∗ ∗ |S11 − S22 |2 = (S11 − S22 )(S11 − S22 ∗ )

= |S12 S21 |2 + (1 − |S22 |2 )(|S11 |2 − ||2 ) Therefore, |1 + |S11 |2 − |S22 |2 − ||2 |2 > 4|S12 S21 |2 + 4(1 − |S22 |2 )(|S11 |2 − ||2 ) or |1 + |S11 |2 − |S22 |2 − ||2 |2 − 4(1 − |S22 |2 )(|S11 |2 − ||2 ) > 4|S12 S21 |2 or (1 − |S22 |2 − |S11 |2 + ||2 )2 > 4|S12 S21 |2 or |1 − |S22 |2 − |S11 |2 + ||2 | > 2|S12 S21 | Therefore,

|1 − |S22 |2 − |S11 |2 + ||2 | >1⇒k>1 2|S12 S21 |

Also, it can be proved that || < 1 implies that B1 > 0 and B2 > 0. Therefore, minus signs must be used in equations (11.2.5) and (11.2.6). Since GT max =

(1 − |MS |2 )|S21 |2 (1 − |ML |2 ) |(1 − S11 MS )(1 − S22 ML ) − S12 S21 ML MS |2

substituting equations (11.2.5), (11.2.6), and k from (11.1.18) gives GT max =

|S21 | (k − k 2 − 1) |S12 |

(11.2.7)

435

AMPLIFIER DESIGN FOR MAXIMUM GAIN

Maximum stable gain is defined as the value of GT max when k = 1. Therefore, S21 GMSG = S12

(11.2.8)

Example 11.3 The S-parameters of a properly biased GaAs FET HFET-1101 are measured using a 50- network analyzer at 6 GHz as follows: S11 = 0.614 −167.4◦ , S21 = 2.187 32.4◦ , S12 = 0.046 65◦ , and S22 = 0.716 −83◦ . Design an amplifier using this transistor for a maximum possible gain. SOLUTION First we need to test for stability using (11.1.16) and (11.1.18). || = |S11 S22 − S12 S21 | = 0.3419 and k=

1 − |S11 |2 − |S22 |2 + ||2 = 1.1296 2|S12 S21 |

Since both of the conditions are satisfied, the transistor is unconditionally stable. For the maximum possible gain, we need to use simultaneous conjugate matching. Therefore, the source and load reflection coefficients are determined from (11.2.5) and (11.2.6) as follows:

MS =

B1 −

2C1

and ML =

B12 − 4|C1 |2

B2 −

B22 − 4|C2 |2 2C2

◦

= 0.8673 169.76

◦

= 0.9011 84.48

There are many ways to synthesize these circuits. The design of one of the circuits is discussed here. As illustrated in Figure 11.7, normalized impedance points A and C are identified corresponding to MS and ML , respectively. The respective normalized admittance points B and D are located next. One way to transform the normalized admittance of unity to that of point B is to add a shunt susceptance (normalized) of about j 2.7 at point E. This can be achieved with a capacitor or a stub of 0.194λ, as shown in Figure 11.8. Now, for moving from point E to point B, we require a transmission line of about 0.065λ. Similarly, an open-circuited shunt stub of 0.297λ will transform the unity value to 1 −j 3.4 at point F, and a transmission line length of 0.093λ will transform it to the desired value at point D. Note that both of the shunt stubs are asymmetrical about the main line. A symmetrical stub is preferable. It can be achieved via two shunt-connected stubs of susceptance j 1.35 at the input end and of −j 1.7 at the output. This circuit

436

TRANSISTOR AMPLIFIER DESIGN

110

0 12

90

100

80

70 60

C

50

30

15

30

0

40

14 0

1

160

10

170

20

E

A 5

2

1

0.5

10

−10

70 0 −1

0.2

0

180

0

B

−

0

−4

0 14

−1

−

50

−16

−20 30

F

−5

30

0

−6

0

−70

−80

−90

20

0

−100 − 11

−1

−1

Figure 11.7

D

Design of input- and output-side matching networks.

0.093λ

0.065λ 50 Ω

0.297λ 50 Ω O.C. 0.194λ

O.C.

Figure 11.8

RF part of the amplifier circuit for Example 11.3.

437

AMPLIFIER DESIGN FOR MAXIMUM GAIN O.C. 0.333λ 0.093λ

O.C. 0.15λ 0.065λ

50 Ω

0.333λ 50 Ω O.C.

0.15λ

O.C.

Figure 11.9

RF part of the amplifier circuit with symmetrical stubs.

is illustrated in Figure 11.9. The value of the gain is evaluated from (11.2.7) as follows: GT max =

|S21 | (k − k 2 − 1) = 28.728 ⇒ 10 log(28.728) = 14.58 dB |S12 |

Unilateral Figure of Merit In preceding examples we found that the design with a bilateral transistor is a bit complex compared with a unilateral case. The procedure for the bilateral transistor becomes even more cumbersome when the specified gain is less than its maximum possible value. It can be simplified by assuming that a bilateral transistor is unilateral. The unilateral figure of merit provides an estimate of the error associated with this assumption. It can be formulated from (10.4.8) and (10.4.10) as follows: 2 GT |1 − S22 L |2 1 − S22 L = = 2 GT U |1 − out L | 1 − S22 L − S12 S21 S L /(1 − S11 S ) 2 1 |1 − S22 L |2 = 1 − S S /(1 − S )(1 − S ) |1 − out L |2 12 21 S L 22 L 11 S or

1 2 GT = GT U 1 − X

where X=

S12 S21 S L (1 − S22 L )(1 − S11 S )

438

TRANSISTOR AMPLIFIER DESIGN

Therefore, the bounds of this gain ratio are given by 1 GT 1 < < 2 (1 + |X|) GT U (1 − |X|)2 ∗ ∗ When S = S11 and L = S22 , GT U has a maximum value. In this case, the maximum error introduced by using GT U in place of GT ranges as follows:

GT 1 1 < < (1 + U )2 GT U (1 − U )2 where U=

|S12 ||S21 ||S11 ||S22 | (1 − |S11 |2 )(1 − |S22 |2 )

(11.2.9)

(11.2.10)

The parameter U is known as the unilateral figure of merit. Example 11.4 The scattering parameters of two transistors are as given below. Compare the unilateral figures of merit of the two. For transistor A, S11 = 0.45 150◦ , S12 = 0.01 −10◦ , S21 = 2.05 10◦ , and S22 = 0.4 −150◦ . For transistor B, S11 = 0.641 −171.3◦ , S12 = 0.057 16.3◦ , S21 = 2.058 28.5◦ , and S22 = 0.572 −95.7◦ . SOLUTION From (11.2.10) we find that U=

0.01 × 2.05 × 0.45 × 0.4 = 0.00551 = UA (1 − 0.452 )(1 − 0.42 )

for transistor A. Similarly, for transistor B, U=

0.057 × 2.058 × 0.641 × 0.572 = 0.1085 = UB (1 − 0.6412 )(1 − 0.5722 )

Hence, the error bounds for these two transistors can be determined from (11.2.9) as follows. For transistor A, 0.9891

0 vL < 0

(13.3.8)

This s(t) can be expressed in terms of the Fourier series as ∞

s(t) =

1 2 nπ + sin cos nωL t 2 n=1 nπ 2

(13.3.9)

For vi = V cos ωi t, the output voltage is found to be ∞ V V 1 nπ cos ωi t + sin {cos[(nωL + ωi )t] + cos[(nωL − ωi )t]} 2 π n=1 n 2 (13.3.10) Thus, the output of this mixer circuit does not contain the local oscillator signal. However, it includes the input signal frequency ωi that appears with the sidebands and spurious signals. It can be a problem in certain applications. For example, consider a case where the input and the local oscillator signal frequencies at a receiver are 60 and 90 MHz, respectively. As illustrated in Figure 13.24, the difference signal frequency at its output is 30 MHz. If a 30-MHz signal is also

vo =

563

SWITCHING-TYPE MIXERS

60 MHz

30 MHz

Filter and IF section

90 MHz

Figure 13.24 Single-balanced mixer of Figure 13.22 at the receiver.

present at the input, it will get through the mixer as well. A preselector (bandpass filter) may be used to suppress the undesired input. Note from (13.3.10) that the amplitude of the sidebands is only V /π, whereas it was twice this in (13.3.6). It influences the sensitivity of the receiver employing this mixer. In the following section we consider another mixer circuit, which solves most of these problems. Double-Balanced Mixer A double-balanced mixer circuit requires four diodes and two transformers with secondary sides center-tapped, as shown in Figure 13.25. As before, it is assumed that the local oscillator voltage vL is high compared with the input vi . Hence, diodes D1 and D2 conduct when the local oscillator voltage is positive while the other two are in the off state. This situation turns the other way around when the voltage vL reverses its polarity. Thus, diodes D1 and D2 conduct for the positive half cycle of vL and D3 and D4 conduct for the negative half. Equivalent circuits can be drawn separately for these two states. Figure 13.26 shows one such equivalent circuit, which represents the case of positive vL . Diode resistance in the forward-biased condition is assumed to be rd . The equivalent circuit can be simplified further to facilitate the analysis. Consider the simplified equivalent circuit shown in Figure 13.27. The currents in the two loops are i1 and i2 . Using Kirchhoff’s voltage law, loop equations can

D4

R

RG + vG −

+ vi −

+ vi − + vi −

D2

+ vL

D3

− D1 RL

Figure 13.25

vo

Switching-type double-balanced mixer.

Local oscillator

564

DETECTORS AND MIXERS R

RG +

+ vi −

vG −

rd

+ vi − + vi −

+ Local oscillator

vL −

rd RL

vo

Figure 13.26 Equivalent of the double-balanced mixer with positive vL . rd + −

vi

+

vL

i1

−

RL

+

i2

vL − rd

Figure 13.27 Simplified equivalent circuit of Figure 13.26.

be written as vi = RL (i1 − i2 ) + rd i1 − vL

(13.3.11)

vi = RL (i1 − i2 ) − rd i2 + vL

(13.3.12)

and

Equations (13.3.11) and (13.3.12) can be solved for i1 − i2 . Hence, i1 − i2 =

vi RL + rd /2

But from Ohm’s law, i1 − i2 = −

vo RL

(13.3.13)

(13.3.14)

Combining (13.3.13) and (13.3.14) yields vo RL =− vi RL + rd /2

(13.3.15)

CONVERSION LOSS

565

Similarly, an equivalent circuit can be drawn for the negative half cycle of vL . Diodes D3 and D4 are conducting in this case, and following an identical analysis it may easily be found that RL vo =+ vi RL + rd /2

(13.3.16)

Equations (13.3.15) and (13.3.16) can be expressed by a single equation after using the function s(t) as defined in (13.3.3). Hence, vo =

RL RL vi = vi s(t) RL + rd /2 RL + rd /2

(13.3.17)

Now using the Fourier series representation for s(t) as given in (13.3.4), (13.3.17) can be expressed as ∞

RL 4 (−1)n+1 nπ vo = vi sin cos nωL t RL + rd /2 π n=1 n 2

(13.3.18)

In case of a sinusoidal input, vi = V cos ωi t, the output voltage vo can be found as vo =

∞ RL 2V (−1)n+1 nπ sin {cos[(nωL + ωi )t] RL + (rd /2) π n=1 n 2

+ cos[(nωL − ωi )t]}

(13.3.19)

Hence, this mixer circuit produces the upper and lower sidebands along with an infinite number of spurious signals. However, the ωL and ωi signals are both isolated from the output. This analysis assumes that the diodes are perfectly matched and the transformers are ideal. Variations of these characteristics may not perfectly isolate the output from the input and the local oscillator. These double-balanced mixers are commercially available for applications in the audio through microwave frequency bands.

13.4 CONVERSION LOSS Since mixers are used to convert the frequency of an input signal, a circuit designer would like to know if there is some change in its power as well. This information is especially important in receiver design where the signal received may be fairly weak. Conversion loss (or gain) of the mixer is defined as a ratio of the power output in one of the sidebands to the power of its input signal. It can be evaluated as follows. Consider the double-balanced mixer circuit shown in Figure 13.24. Using its equivalent circuit illustrated in Figure 13.26, the input resistance Rin can be found

566

DETECTORS AND MIXERS

with the help of (13.3.13) as follows: Rin =

vi rd ≈ RL = RL + i1 − i2 2

(13.4.1)

It is assumed here that load resistance RL is much larger than the forward resistance rd of the diode. If there is maximum power transfer from the voltage source vG at the input to the circuit, the source resistance RG must be equal to the input resistance Rin . Since the input resistance is equal to the load RL , the following condition must be satisfied: RG = RL If the sinusoidal source voltage vG has a peak value of VP , it divides between two equal resistances RG and Rin , and only half of voltage VP appears across the transformer input. Hence, V2 Pi = P (13.4.2) 8RL From (13.3.19), the peak output voltage Vo of the sideband is Vo |ωL ±ωi =

2V VP = π π

(13.4.3)

Therefore, the output power Po is found as Po =

VP2 2π2 RL

(13.4.4)

If G is defined as the conversion gain, which is a ratio of its output to input power, then from (13.4.2) and (13.4.4) we find that G=

Po VP2 8RL 4 = = 2 Pi 2π2 RL VP2 π

(13.4.5)

Since G is less than unity, it is really not a gain but a loss of power. Hence, the conversion loss, L, is L = 10 log

π2 = 3.92 dB ≈ 4 dB 4

(13.4.6)

Hence, approximately 40% of the power of the input signal is transferred to the output of an ideal double-balanced mixer. Circuit loss and mismatch reduce it further. In the case of a single-balanced mixer where the local oscillator is isolated from the output, the sideband voltage can be found from (13.3.10). Hence, the

INTERMODULATION DISTORTION IN DIODE-RING MIXERS

power output in one of the sidebands is given as 2 1 VP2 V Po = = 2RL π 8π2 RL

567

(13.4.7)

Since the input power is still given by (13.4.2), its conversion gain may be found as Vs2 8RL 1 Po = × 2 = 2 (13.4.8) G= 2 Pi 8π RL Vs π Again, the output is less than the input power, and therefore, it represents a loss of signal power. The conversion loss for this case is L = 10 log π2 = 9.943 ≈ 10 dB

(13.4.9)

Hence, the conversion loss in this single-balanced mixer is 6 dB (four times) larger than that of the double-balanced mixer considered earlier. It may be noted that the mixer circuit shown in Figure 13.19 has a conversion loss of only 4 dB. These results show that the maximum possible power transferred from the input to the output of a diode mixer is less than 40%. Therefore, these circuits always have conversion loss. A conversion gain is possible only when a transistor (BJT or FET) circuit is used as the mixer. FET mixers are described briefly later in the chapter.

13.5 INTERMODULATION DISTORTION IN DIODE-RING MIXERS As discussed in Chapter 2, the noise level defines the lower end of the dynamic range of a system, whereas the distortion determines its upper end. Therefore, a circuit designer needs to know the distortion introduced by the mixer and whether there is some way to limit that. We consider here the mixer circuit depicted in Figure 13.28, which has a resistance R in series with each diode. D4

R

RG +

+ vi −

vG −

+ vi −

D2

R

+ vL

D3 R

+ vi −

R

v

−

D1 RL

vo

Figure 13.28 Mixer circuit with resistance R in each branch.

Local oscillator

568

DETECTORS AND MIXERS D1

R

Ro

iL − i + vL −

+ vo − iL + i

+ vi −

D2

RL

v

R

Figure 13.29 Simplified equivalent of the circuit shown in Figure 13.28 for vL > 0.

Assuming that vL is large enough that the local oscillator alone controls the conduction through the diodes, an equivalent circuit can be drawn for each half cycle. In the case of a positive half period of vL , the simplified equivalent circuit may be drawn as illustrated in Figure 13.29. Diodes D1 and D2 conduct for this duration while the other two are in the off state. Assume that iL is current due to vL and i is due to vi . Hence, currents through the diodes are given as follows: current through diode D1 = id3 = iL − i current through diode D2 = id4 = iL + i Since the V –I characteristic of a diode is given as id = Is (eαvd − 1) ≈ Is eαvd where α= we find that vd =

(13.5.1)

q nkT

(13.5.2)

1 id ln α Is

(13.5.3)

Note that net current flowing through the load resistance RL is the difference of two loop currents: iL − i and iL + i. Hence, the LO current through RL cancels out, and two components of current due to vi are added to make it 2i. Loop equations for the circuit can be found to be vL = (iL − i)R + vd3 + vi − RL · 2i

(13.5.4)

vL = RL · 2i − vi + R(iL + i) + vd4

(13.5.5)

and

Subtracting (13.5.5) from (13.5.4), we have 0 = −2(R + 2RL )i + 2vi + vd3 − vd4

569

INTERMODULATION DISTORTION IN DIODE-RING MIXERS

or

1 iL + i 1 iL + i iL − i 2(R + 2RL )i − 2vi = − ln = − ln ln α Is Is α iL − i

or vi = (R + 2RL )i +

1 iL + i ln 2α iL − i

(13.5.6)

Since the LO current iL is very large compared with i, it can be expanded as an infinite series: 1 1 5 1 3 i + i + ··· (13.5.7) vi = R + 2RL + i+ 3 αiL 3αiL 5αiL5 An inverse series solution to this can easily be found with the help of computer software (such as Mathematica) as follows: i=

vi3 vi − + ··· R + 2RL + 1/αiL 3αiL3 (R + 2RL + 1/αiL )4

(13.5.8)

Therefore, the output voltage vo can be written as vi3 vi vo = 2RL i = 2RL − + ··· R + 2RL + 1/αiL 3αiL3 (R + 2RL + 1/αiL )4 (13.5.9) On comparing it with (2.5.31), we find that the term responsible for the intermodulation distortion (IMD) is

k3 = −

3αiL3

2RL (R + 2RL + 1/αiL )4

(13.5.10)

Since the IMD is directly proportional to the cube of k3 , (13.5.10) must be minimized to reduce the intermodulation distortion. One way to achieve this objective is to use a large local oscillator current iL . Another possible way is to employ the resistance R in series with each diode. Sometimes an additional diode is used in place of R. This permits higher local oscillator drive levels and a corresponding reduction in IMD. Example 13.4 In the mixer circuit shown in Figure 13.30, transformer and diodes are ideal. The transformer has 400 turns on its primary and 800 turns with a center tap on its secondary side. Find the output voltage and the value of load RL that provides maximum power transfer from vs . SOLUTION For maximum power transfer, the impedance when transformed to the primary side of the transformer must be equal to 50 . Since the number of turns on either side of the center tap is the same as on its primary side and only one of the diodes is conducting at a given time, the load resistance RL must be

570

DETECTORS AND MIXERS

50 Ω

RL

vi vs

vo

vL

vi

RL

Figure 13.30 Mixer circuit for Example 13.4.

50 as well. Under this condition, only half of the signal voltage appears across the primary. Hence, vi = vs /2. Assume that vL is large compared with vi such that diode switching is controlled solely by the local oscillator. Therefore, v + vi vo = L −(vL + vi )

for vL > 0 for vL < 0

This output voltage can be expressed as follows: vo = (vL + vi )s(t) where

1 vL > 0 s(t) = −1 vL < 0

We already know the Fourier series representation for this s(t), given by (13.3.4) as ∞ nπ 4 (−1)n+1 sin cos nωL t s(t) = (13.3.4) π n=1 n 2 If vi and vs are sinusoidal voltages vi = Vi cos ωi t and vL = VL cos ωL t then the output voltage will be ∞

vo = (Vi cos ωi t + VL cos ωL t)

nπ 4 (−1)n+1 sin cos nωL t π n=1 n 2

571

FET MIXERS

or ∞

vo =

2 (−1)n+1 nπ sin {Vi [cos(nωL + ωi )t + cos(nωL − ωi )t] π n=1 n 2 + VL [cos(nωL + ωL )t + cos(nωL − ωL )t]}

13.6 FET MIXERS Diodes are commonly used in mixer circuits because of certain advantages, including a relatively low noise. A major disadvantage of these circuits is the conversion loss, which cannot be reduced below 4 dB. On the other hand, transistor circuits can provide a conversion gain, although they may be noisier. A well-designed FET mixer produces less distortion than do BJT circuits. Further, the range of its input voltage is generally much higher (10 times or so) with respect to BJT mixers. The switching and resistive types of mixer have been designed using the FET. In this section we present an overview of selected FET mixer circuits. Figure 13.31 shows a FET mixer circuit. Local oscillator signal vL is applied to the source terminal of the FET via capacitor C1 while the input v2 is connected to its gate. For the FET operating in saturation, we can write Vgs 2 (13.6.1) iD = IDSS 1 − Vp where iD is the drain current of the FET, Vgs the gate-to-source voltage, Vp its pinch-off voltage, and IDSS the drain current for Vgs = 0. For the circuit shown in Figure 13.31, the total gate voltage Vgs is Vgs = vi − vL + VGS VDD vo

C2

C1 vi

RG

RS vL

Figure 13.31

FET mixer circuit.

572

DETECTORS AND MIXERS

where VGS is the gate-to-source dc bias voltage. Therefore, the drain current iD in this case is vi − vL + VGS 2 iD = IDSS 1 − (13.6.2) Vp For sinusoidal vi and vL given as vi = Vi cos ωi t and vL = VL cos ωL t the drain current iD can be found from (13.6.2). Hence, Vi cos ωi t − VL cos ωL t + VGS 2 iD = IDSS 1 − Vp or 2 [Vi cos ωi t − VL cos ωL t + VGS ] iD = IDSS 1 − Vp +

IDSS 2 [V + 2VGS Vi cos ωi t + Vi2 cos2 ωi t − 2VL VGS cos ωL t Vp2 GS + VL2 cos2 ωL t − 2Vi VL cos ωi t cos ωL t]

(13.6.3)

Equation (13.6.3) can be rearranged in the following form and then the amplitudes of sum and difference frequency components can easily be identified as follows: iD = IDC + a1 cos ωi t + a2 cos ωL t + b1 cos 2 ωi t + 2 cos 2 ωL t − c{cos[(ωL + ωi )t] + cos[(ωL − ωi )t]} where c=

IDSS Vi VL Vp2

(13.6.4)

The ratio of amplitude c to input voltage Vi is defined as the conversion transconductance gc . Hence, c IDSS VL = (13.6.5) gc = Vi Vp2 For high conversion gain, gc must be large. It appears from (13.6.5) that the FET with high IDSS should be preferable, but that is generally not the case. IDSS and Vp of a FET are related such that the device with high IDSS also has high Vp . Therefore, it may result in a lower gc than that of a low IDSS device. VL

573

FET MIXERS

is directly related to the conversion gain. Hence, higher local oscillator voltage increases the conversion gain. Since the FET is to be operated in its saturation mode (the constant-current region), VL must be less than the magnitude of the pinch-off voltage. As a special case, if VL = |Vp |/2, the conversion transconductance may be found as IDSS gc = (13.6.6) 2Vp Since the transconductance gm of a JFET is given as gm =

∂iD 2IDSS =− ∂VGS Vp

1−

VGS Vp

(13.6.7)

and gm |VGS =0 = −

2IDSS Vp

(13.6.8)

the conversion transconductance is one-fourth of the small-signal transconductance evaluated at VGS = 0 (provided that VL = 0.5Vp ). For a MOSFET, it can be shown that the conversion transconductance cannot exceed one-half of the small-signal transconductance. Note from (13.6.3) that there are fundamental terms of input as well as local oscillator frequencies, their second-order harmonics, and the desired ωi ± ωL components. Unlike the diode mixer, higher-order spurious signals are not present in this output. In reality, there is some possibility for those terms to be present because of the circuit imperfections. An alternative to the circuit shown in Figure 13.31 combines the local oscillator signal with the input before applying it to the gate of FET. The analysis of this circuit is similar to the one presented above. In either case, one of the serious problems is isolation among the three signals. There have been numerous attempts to address those problems. Example 13.5 A given JFET has IDSS = 50 mA and transconductance gm = 200 mS when its gate voltage VGS is zero. If it is being used in a mixer that has a 50- load, find the conversion gain of the circuit. SOLUTION Since gm |VGS =0 = − then Vp =

2IDSS Vp

2 × 50 × 10−3 1 = = 0.5 V 200 × 10−3 2

574

DETECTORS AND MIXERS

If the local oscillator voltage VL is kept at approximately 0.25 V, the conversion transconductance gc may be found as gc =

50 × 10−3 = 50 × 10−3 S = 50 mS 2 × 0.5

Therefore, the magnitude of voltage gain Av is Av ≈ gc RL = 50 × 10−3 × 50 = 2.5 Since a common-gate configuration of the FET is generally used in the mixer circuit, its current gain is close to unity. Therefore, the conversion power gain of this circuit is approximately 2.5. A dual-gate FET is frequently employed in a mixer circuit. There are several circuit configurations and modes of operation for dual-gate FET mixers. Figure 13.32 illustrates a circuit that usually performs well in most receiver designs. The working of this circuit can be explained considering that a dualgate FET represents two single-gate transistors connected in cascode. Hence, the upper FET works as a source follower and the lower one provides mixing. An RF signal is applied to a gate that is close to the ground while the local oscillator is connected at the other gate (one close to the drain in common-source configuration). If RF is connected to the other gate, the drain resistance of the lower transistor section appears as source resistance and thus reduces the signal. This circuit is used as a mixer when both transistors are operating in their triode (nonsaturation) mode. A bypass circuit is needed before its output to block the input (RF) and the local oscillator signals. Further, another bypass circuit may be needed across the local oscillator to block the output frequencies. This circuit has lower gain and higher noise, but a properly designed mixer can achieve a modestly higher third-harmonic intercept compared with a single-gate mixer. Figure 13.33 shows a FET mixer circuit that employs two transistors and transformers. A local oscillator signal is applied to both transistors in the same phase while one RF input is out of phase with respect to the other. The resulting output is a differential of the two sides, and therefore the local oscillator signal is

Local oscillator vL

Matching network

RF input vi

Matching network

Matching network

Figure 13.32 Dual-gate FET mixer.

Output vo

575

FET MIXERS

vi

vo

+ VDD

vL

RL

Rs

Figure 13.33 FET balanced mixer.

canceled out while the RF is added. The nonlinear characteristic of the transistors generates the upper and lower sideband signals. If the dc bias voltage at the source is VS , the RF input is Vi cos ωi t, and the local oscillator voltage is VL cos ωL t, the output voltage of this circuit is vo =

4RL Vi 4RL Vi VS + VP VP2

cos ωi t +

2RL Vi VL VP2

× [cos(ωL + ωi )t + cos(ωL − ωi )t]

(13.6.9)

This shows that the output of this circuit contains the RF input frequency along with two sidebands. Asymmetry and other imperfection can add more frequency components as well. Figure 13.34 illustrates an active double-balanced circuit, known as the Gilbert cell mixer, commonly employed in radio-frequency integrated circuits (RFICs). Transistors Q1 and Q2 are used to convert the RF voltage to current, and Q3 to Q6 are used for the switching. RF and local oscillator signals are applied to the circuit via the corresponding matching networks (baluns). Similarly, the IF output requires a matching network as well. A detailed analysis of the circuit is beyond the scope of this book; only a qualitative analysis is presented here. Assume that the local oscillator signal is a large square wave that can turn on the transistors when it is high and turn them off when it is low. Note that the current iif1 is the sum of i3 and i5 , and iif2 is the sum of i4 and i6 . Further, the RF current i1 is the sum of i3 and i4 and the sum of i5 and i6 is equal to i2 . The bias current Ib is the sum of i1 and i2 . Since a transistor is on only when the local oscillator signal is high, transistors Q4 and Q5 will be on while Q3 and Q6 will be in off conditions. Therefore, iif1 = i5 = i2 and iif2 = i4 = i1

576

DETECTORS AND MIXERS

VDD RD RD iif2

iif1

Balun

LO

i3 Q3

Balun i5

i4 Q4

Q6

i2

Q1 RF

i6

Q5

i1

IF

Q2

Balun Ib

Figure 13.34 Gilbert cell mixer circuit.

Similarly, when the local oscillator signal is low, transistors Q3 and Q6 will be on and Q4 and Q5 will be off. Therefore, iif1 = i3 = i1 and iif2 = i6 = i2 Hence the IF output vif can be found to be vif =

RD (i2 − i1 ) RD (i1 − i2 )

vL > 0 vL < 0

(13.6.10)

or vif = RD (i2 − i1 )s(t)

(13.6.11)

where s(t) is defined by (13.3.3) and (13.3.4). Further, the conversion gain Gc of this circuit can be found after assuming that it is symmetrical, and therefore the transconductance gm of Q1 is the same as that of Q2 . It is given as Gc =

4 gm RD π

(13.6.12)

577

PROBLEMS

SUGGESTED READING Bahl, I. J., and P. Bhartia, Microwave Solid State Circuit Design. New York: Wiley, 1988. Collin, R. E., Foundations for Microwave Engineering. New York: McGraw-Hill, 1992. Larson, L. E. (ed.), RF and Microwave Circuit Design for Wireless Communications. Boston: Artech House, 1996. Leung, B., VLSI for Wireless Communication. Upper Saddle River, NJ: Prentice Hall, 2002. Maas, S. A., Microwave Mixers. Boston: Artech House, 1993. Pozar, D. M., Microwave Engineering. New York: Wiley, 1998. Rohde, U. L., Microwave and Wireless Synthesizers. New York: Wiley, 1997. Smith, J. R., Modern Communication Circuits. New York: McGraw-Hill, 1998. Vendelin, G. D., A. Pavio, and U. L. Rhode, Microwave Circuit Design Using Linear and Non-linear Techniques. New York: Wiley, 1990.

PROBLEMS 13.1. In an AM signal, the carrier output is 1 kW. If the wave modulation is 100%, determine the power in each sideband. How much power is being transmitted? 13.2. Assume that an AM transmitter is modulated with a video signal given by m(t) = −0.15 + 0.7 sin ω1 t, where f1 is 4 MHz. Let the unmodulated carrier amplitude be 100 V. Evaluate and sketch the spectrum of modulated signal. 13.3. An AM broadcast station that uses 95% modulation is operating at total output power of 100 kW. Find the power transmitted in its sidebands. 13.4. With a modulation index of 80% an AM transmitter produces 15 kW. How much of this is carrier power? Find the percentage power saving if the carrier and one of the sidebands are suppressed before transmission. 13.5. When an AM station transmits an unmodulated carrier, the current through its antenna is found to be 10 A. It increases to 12 A with a modulated signal. Determine the modulation index used by the station. 13.6. In the circuit arrangement shown in Figure P13.6, Vout = V1 − V2 is the output signal, where V1 = 10VC + 0.2VC Vm and V2 = 6VC − 0.2VC Vm .

Vout VC = C cos ωct

V1

V2 Vm = M cos ωmt

Figure P13.6

578

DETECTORS AND MIXERS

(a) Calculate the value of M that results in a modulation index of 0.8. (b) Determine the value of C that results in an unmodulated output voltage amplitude of 8 V. (c) If this Vout is applied to a 50- load, find the power in the carrier and in each sideband. 13.7. The circuit shown in Figure P13.7 uses a zero-biased diode with Rj = 500 , and n = 1 at 290 K. If the signal applied to this circuit is vin (t) = 0.5[1 + 0.3 cos(4π × 104 t + 0.15)] cos(π × 109 t − 0.15)

V

find the output voltage vout . Also, sketch the frequency spectra of vin and vout .

1 kΩ

31.83 pF vout

vin

Figure P13.7

13.8. A 1-MHz carrier signal is amplitude-modulated simultaneously with 300Hz, 800-Hz, and 2-kHz sinusoidal signals. What frequencies are present in the output? 13.9. An AM broadcast transmitter radiates 50 kW of carrier power. If the modulation index is 0.85, find the power radiated. 13.10. An FM broadcast station uses a 10-kHz audio signal of 2 V peak value to modulate the carrier. If the allowed frequency deviation is 20 kHz, determine the bandwidth requirement. 13.11. A frequency modulator is supplied with a carrier of 10 MHz and a 1.5kHz audio signal of 1 V. The frequency deviation at its output is found to be ±9 kHz. (a) An FM wave from the modulator is passed through a series of frequency multipliers with a total multiplication of 15 (i.e., fo = 15fc ). Determine the frequency deviation at the output of the multiplier. (b) An output of the modulator is mixed with a 58-MHz signal. Find the sum frequency output of the mixer and its frequency deviation. (c) Find the modulation index at the output of the modulator. (d) The audio input to the modulator is changed to 500 Hz with its amplitude at 2 V. Determine the modulation index and frequency deviation at the output of the modulator. 13.12. An FM signal, vFM (t) = 1500 cos[2π × 109 t + 2 sin(3π × 105 )]

V

579

PROBLEMS

is applied to a 75- antenna. Determine (a) the carrier frequency, (b) the transmitted power, (c) the modulation index, and (d) the frequency of the modulating signal. 13.13. When the modulating frequency in FM is 400 Hz and the modulating voltage is 2.4 V, the modulation index is 60. Calculate the maximum deviation in frequency. What is the modulation index when the modulating frequency is reduced to 250 Hz and the modulating voltage is simultaneously raised to 3.2 V? 13.14. The iD –Vgs characteristics of a FET are given by

2Vgs iD = 7 1 − 7

2 mA

If it is being used in a mixer that has a 50- load, find the conversion gain of the circuit. 13.15. Calculate the conversion loss of the double-balanced mixer shown in Figure P13.15. The diode ON resistance is much less than the load resistance RL . What value of RL is required for maximum power transfer if the transformer employed in the circuit has 400 turns in its primary and 800 turns with a center tap on its secondary side? 50 Ω

vi vs

vi

RL vo

vL RL

Figure P13.15

13.16. In the mixer circuit shown in Figure P13.16, the transformer and diodes are ideal. The transformer has 200 turns on each side. Find the value of RL that provides maximum power transfer from the source. 75 Ω

vi vs

vi

RL vo

vL RL

Figure P13.16

APPENDIX 1 DECIBELS AND NEPER

Consider the two-port network shown in Figure A1.1. Assume that V1 and V2 are the voltages at its ports. The voltage gain (or loss) Gv of this circuit is expressed in decibels as V2 Gv = 20 log10 dB (A1.1) V1 Similarly, if P1 and P2 are the power levels at the two ports, the power gain (or loss) of this circuit in decibels is given as Gp = 10 log10

P2 P1

dB

(A1.2)

Thus, the dB unit provides a relative level of the signal. For example, if we are asked to find P2 in watts for Gp as 3 dB, we also need P1 . Otherwise, the only information we can deduce is that P2 is twice P1 . Sometimes power is expressed in logarithmic units, such as dBW and dBm. These units are defined as follows. If P is power in watts, it can be expressed in dBW as G = 10 log10 P dBW (A1.3) On the other hand, if P is in milliwatts, the corresponding dBm power is found as G = 10 log10 P

dBm

(A1.4)

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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581

DECIBELS AND NEPER

Two-port network

V1

V2

Figure A1.1 Two-port network.

Thus, the dBW and dBm units represent power relative to 1 W and 1 mW, respectively. Another decibel unit that is commonly used to specify phase noise of an oscillator or strengths of various sidebands of a modulated signal is dBc. It specifies the signal strength relative to the carrier. Consider a 100-MHz oscillator that has an output power of −10 dBm. Suppose that its output power is −30 dBm in the frequency range 105 to 106 MHz; then power per hertz in the output spectrum is −90 dBm and the phase noise is −80 dBc. In general, if amplitudes of the sideband and the carrier are given as V2 and Vc , respectively, the sideband in dBc is found as G2 = 20 log

V2 Vc

(A1.5)

Consider now a 1-m-long transmission line with its attenuation constant α. If V1 is the signal voltage at its input port, the voltage V2 at its output is given as |V2 | = |V1 |e−α

(A1.6)

Therefore, the voltage gain Gv of this circuit in nepers is Gv (NP ) = ln

|V2 | = −α |V1 |

NP

(A1.7)

On the other hand, Gv in decibels is found to be Gv (dB) = 20 log

|V2 | = 20 log e−α = −20α log e = −8.6859α |V1 |

dB

Therefore, 1NP = 8.6859 dB

(A1.8)

The negative sign indicates that V2 is smaller than V1 and there is loss of signal.

APPENDIX 2 CHARACTERISTICS OF SELECTED TRANSMISSION LINES

COAXIAL LINE Consider a coaxial line with its inner and outer conductor radii a and b, respectively, as illustrated in Figure A2.1. Further, εr is the dielectric constant of the insulating material. The line parameters L, R, C, and G of this coaxial line are found as follows: 55.63εr C= pF/m (A2.1) ln(b/a) and L = 200 ln

b a

nH/m

(A2.2)

If the coaxial line has small losses due to imperfect conductor and insulator, its resistance and conductance parameters can be calculated as follows:

1 1 R ≈ 10 + a b

fGHz σ

/m

(A2.3)

and G=

0.3495εr fGHz tan δ ln(b/a)

S/m

(A2.4)

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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CHARACTERISTICS OF SELECTED TRANSMISSION LINES

583

a

εr

b

Figure A2.1 Coaxial line geometry.

where tan δ is the loss tangent of the dielectric material, σ the conductivity of the conductors in S/m, and fGHz the signal frequency in GHz. The characteristic impedance and propagation constant of the coaxial line can be calculated easily using the formulas given in Chapter 3. Attenuation constants αc and αd due to conductor and dielectric losses, respectively, may be determined from Rs 1 1 αc = √ + (A2.5) b 2 µ0 /ε0 εr ln(b/a) a ω√ µ0 ε0 εr tan δ (A2.6) αd = 2 where ωµ0 Rs = (A2.7) 2σ STRIP LINE Strip line geometry is illustrated in Figure A2.2. Insulating material of thickness h has a dielectric constant εr . The width and thickness of the central conducting strip are w and t, respectively. For the case of t = 0, its characteristic impedance can be found as follows: √ −1 1 + x 296.1 0.6931 + ln 0 < x ≤ 0.7 √ 1 1 − x Z0 = √ (A2.8) √ εr x 1 + 30 0.6931 + ln √ 0.7 ≤ x < 1 1− x πw (A2.9) x = tanh 2h

584

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

w t

h

εr

Figure A2.2 Strip line geometry.

and x=

1 − x 2

(A2.10)

For the design of a strip line, the following convenient formulas can be used: √ w = 0.6366 tanh−1 k h

(A2.11)

where

[(eπ/y − 2)/(eπ/y + 2)]2

1 − [(eπy − 2)/(eπy + 2)]4 √ εr y = Z0 94.18 k=

0≤y≤1 1≤y

(A2.12) (A2.13)

For t = 0, w = 1 − 2 h

(A2.14) √

1 = 2.5465(1 − ) and

ey + 0.568 ey − 1

2 δ 0.0796 1 − 0.5 ln + 2 = π 2− 1 − 0.26

(A2.15)

(A2.16)

where = δ= and y is as defined in (A2.13).

t h

(A2.17) 2

1+

2 [/(1 3

− )]

(A2.18)

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

585

MICROSTRIP LINE The geometry of a microstrip line is illustrated in Figure A2.3. Dielectric substrate on the conducting ground is h meters high and its dielectric constant is εr . The width and thickness of the conducting strip on its top are w and t, respectively. The characteristic impedance of this line can be found as follows: 8h we we 60 for ≤1 √εre ln we + 4h h Z0 = w −1 376.7 we we e + 1.393 + 0.667 ln + 1.444 ≥1 for √ εre h h h (A2.19) where w w w t 1 + 0.3979 1 + ln 12.5664 ≤ h h t h 2π we (A2.20) = w t h w 1 h + 0.3979 1 + ln 2 ≥ h h t h 2π The effective dielectric constant εre of a microstrip line ranges between εr and 1 because of its propagation characteristics. If the signal frequency is low such that the dispersion is not a problem, it can be determined as follows: w ε − 1 t h r − εre = 0.5 εr + 1 + (εr − 1)F (A2.21) h 4.6 h w where w 2 h −0.5 1 + 12 + 0.04 1 − w w h = F −0.5 h h 1 + 12 w

w ≤1 h w ≥1 h

w t

h

εr

Figure A2.3

Microstrip line geometry.

(A2.22)

586

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

If dispersion cannot be ignored, the effective dielectric constant may be found as follows: √ 2 √ εr − εre √ εre (f ) = + εre (A2.23) 1 + 4F −1.5 where

40 w 2 F = fGHz h εr − 1 0.5 + 1 + 2 log 1 + 3 h

(A2.24)

The corresponding characteristic impedance is determined from the following formula: εre (f ) − 1 εre Z0 (f ) = Z0 (A2.25) εre − 1 εre (f ) Attenuation constants αc and αd for the conductor and dielectric losses, tively, are determined as follows: fGHz 32 − (we / h)2 ζ Np/m 9.9825 hZ0 σ 32 + (we / h)2 αc = 0.667(we / h) ζZ0 εre fGHz we −5 44.1255 × 10 + Np/m h σ h (we / h) + 1.444

respec-

w ≤1 h w ≥1 h (A2.26)

where

w 1.25 1.25t h 1 + we 1 + πw + π ln 4π t ζ= h 1.25t 1.25 h 1− + ln 2 1+ we πh π t

w 1 ≤ h 2π w 1 ≥ h 2π

(A2.27)

σ is conductivity of the conductor and fGHz is the signal frequency in GHz. αd = 10.4766

εr εre − 1 fGHz tan δ √ εr − 1 εre

Np/m

(A2.28)

For the design of a microstrip line that has negligible dispersion, the following formulas may be more convenient. For A > 1.52, w 8eA = 2A h e −2

(A2.29)

For A ≤ 1.52,

w 0.61 εr − 1 ln(B − 1) + 0.39 − = 0.6366 B − 1 − ln(2B − 1) + h 2εr εr (A2.30)

CHARACTERISTICS OF SELECTED TRANSMISSION LINES

587

where A=

εr − 1 Z0

0.11 εr + 1 + 0.23 + 84.8528 εr + 1 εr

(A2.31)

B=

592.2 √ Z0 εr

(A2.32)

and

Experimental verification indicates that (A2.29) and (A2.30) are fairly accurate as long as t/ h ≤ 0.005.

APPENDIX 3 SPECIFICATIONS OF SELECTED COAXIAL LINES AND WAVEGUIDES

TABLE A3.1 Cable Type RG(−) 8/U 9/U 11/U 58/U 59/U 122/U 141A/U 142B/U 174/U 178B/U 179B/U 180B/U 213/U 214/U 223/U 316/U

Selected Computer, Instrumentation, and Broadcast Cables

Insulation

Core O.D. (in.)

Cable O.D. (in.)

Zo ()

Capacitance (pF/ft)

Attenuation (dB/100 ft) at 400 MHz

Polyethylene Polyethylene Polyethylene Polyethylene Polyethylene Polyethylene Teflon Teflon Teflon Teflon Teflon Teflon Polyethylene Polyethylene Polyethylene Teflon

0.285 0.280 0.285 0.116 0.146 0.096 0.116 0.116 0.060 0.034 0.063 0.102 0.285 0.285 0.116 0.060

0.405 0.420 0.405 0.195 0.242 0.160 0.190 0.195 0.100 0.072 0.100 0.140 0.405 0.425 0.206 0.098

50 51 75 53.5 75 50 50 50 50 50 75 95 50 50 50 50

26.0 30.0 20.5 28.5 17.3 30.8 29.0 29.0 30.8 29.0 19.5 15.0 30.8 30.8 30.8 29.0

3.8 4.1 4.2 9.5 5.6 16.5 9.0 9.0 20.0 28.0 21.0 17.0 4.7 4.7 10.0 20.0

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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589

SPECIFICATIONS OF SELECTED COAXIAL LINES AND WAVEGUIDES

TABLE A3.2

MIL-C-17F Designation

Selected Semirigid Coaxial Lines Nominal Impedance ()

Dielectric Diameter (in.)

Center Conductor Diameter (in.)

Maximum Operating Frequency (GHz)

Capacitance (pF/ft)

Attenuation at 1 GHz (dB/100 ft)

50 50 50 50 50

0.209 0.1175 0.066 0.037 0.026

0.0641 0.0362 0.0201 0.0113 0.008

18 20 20 20 20

29.6 29.9 32 32 32

7.5 12 22 40 60

129 130 133 151 154

TABLE A3.3

Standard Rectangular Waveguides

EIA Nomenclature WR (−)

Inside Dimension (in.)

TE10 Mode Cutoff Frequency (GHz)

Recommended Frequency Band for TE10 Mode

2300 2100 1800 1500 1150 975 770 650 510 430 340 284 229 187 159 137 112 90 75 62 51 42 34 28 22 19 15 12 10 8 7 5 4 3

23.0 × 11.5 21.0 × 10.5 18.0 × 9.0 15.0 × 7.5 11.5 × 5.75 9.75 × 4.875 7.7 × 3.85 6.5 × 3.25 5.1 × 2.55 4.3 × 2.15 3.4 × 1.7 2.84 × 1.34 2.29 × 1.145 1.87 × 0.872 1.59 × 0.795 1.372 × 0.622 1.122 × 0.497 0.9 × 0.4 0.75 × 0.375 0.622 × 0.311 0.51 × 0.255 0.42 × 0.17 0.34 × 0.17 0.28 × 0.14 0.244 × 0.112 0.188 × 0.094 0.148 × 0.074 0.122 × 0.061 0.1 × 0.05 0.08 × 0.04 0.065 × 0.0325 0.051 × 0.0255 0.043 × 0.0215 0.034 × 0.017

0.2565046 0.2809343 0.3277583 0.3933131 0.5130267 0.6051054 0.7662235 0.9077035 1.1569429 1.3722704 1.7357340 2.0782336 2.5779246 3.1530286 3.7125356 4.3041025 5.2660611 6.5705860 7.8899412 9.4951201 11.586691 14.088529 17.415732 21.184834 26.461666 31.595916 40.058509 48.54910 59.35075 74.44066 91.22728 116.47552 137.93866 174.43849

0.32–0.49 0.35–0.53 0.41–0.62 0.49–0.75 0.64–0.98 0.76–1.15 0.96–1.46 1.14–1.73 1.45–2.2 1.72–2.61 2.17–3.30 2.60–3.95 3.22–4.90 3.94–5.99 4.64–7.05 5.38–8.17 6.57–9.99 8.20–12.5 9.84–15.0 11.9–18.0 14.5–22.0 17.6–26.7 21.7–33.0 26.4–40.0 32.9–50.1 39.2–59.6 49.8–75.8 60.5–91.9 73.8–112 92.2–140 114–173 145–220 172–261 217–330

APPENDIX 4 SOME MATHEMATICAL FORMULAS

sin(A + B) = sin A cos B + cos A sin B sin(A − B) = sin A cos B − cos A sin B cos(A + B) = cos A cos B − sin A sin B cos(A − B) = cos A cos B + sin A sin B tan A + tan B tan(A + B) = 1 − tan A tan B tan A − tan B tan(A − B) = 1 + tan A tan B sin2 A + cos2 A = 1 tan2 A + 1 = sec2 A cot2 A + 1 = csc2 A A−B A+B cos 2 2 A−B A+B sin sin A − sin B = 2 cos 2 2 A+B A−B cos A + cos B = 2 cos cos 2 2 sin A + sin B = 2 sin

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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SOME MATHEMATICAL FORMULAS

A+B B−A sin 2 2 2 sin A cos B = sin(A + B) + sin(A − B) cos A − cos B = 2 sin

2 cos A sin B = sin(A + B) − sin(A − B) 2 cos A cos B = cos(A + B) + cos(A − B) 2 sin A sin B = cos(A − B) − cos(A + B) 1 − cos A A sin = ± 2 2 sin 2A = 2 sin A cos A cos 2A = cos2 A − sin2 A = 2 cos2 A − 1 = 1 − 2 sin2 A 2 tan A tan 2A = 1 − tan2 A A 1 + cos A cos = ± 2 2 A 1 − cos A sin A 1 − cos A tan = ± = = 2 1 + cos A 1 + cos A sin A ej A = cos A + j sin A e−j A = cos A − j sin A sin A =

ej A − e−j A A3 A5 A7 =A− + − + ··· 2j 3! 5! 7!

ej A + e−j A A2 A4 A6 =1− + − + ··· 2 2! 4! 6! A5 A7 A3 e−j A − ej A + 2 + 17 +··· = A + tan A = j j A e + e−j A 3 15 315 cos A =

eA − e−A 2 A e + e−A cosh A = 2 sinh A eA − e−A tanh A = = A cos A e + e−A sinh A =

cosh A eA + e−A = A sinh A e − e−A sinh(A + B) = sinh A cosh B + cosh A sinh B coth A =

cosh(A + B) = cosh A cosh B + sinh A sinh B

591

592

SOME MATHEMATICAL FORMULAS

sinh(A − B) = sinh A cosh B − cosh A sinh B cosh(A − B) = cosh A cosh B − sinh A sinh B tanh A + tanh B tanh(A + B) = 1 + tanh A tanh B tanh A − tanh B tanh(A − B) = 1 − tanh A tanh B cos j A = cosh A sin j A = j sinh A cosh2 A − sinh2 A = 1 sinh j A = j sin A cosh j A = cos A tanh j A = j tan A A3 A4 A2 + + +··· 2! 3! 4 A2 A3 A4 e−A = 1 − A + − + − ··· 2 3 4! A3 A5 A7 eA − e−A =A+ + + + ··· sinh A = 2 3! 5! 7! eA + e−A A2 A4 A6 cosh A = =1+ + + +··· 2 2! 4! 6! loga xy = loga x + loga y x loga = loga x − loga y y eA = 1 + A +

loga x y = y loga x loga x = logb x × loga b =

logb x logb a

ln x = log10 x × ln 10 = 2.302585 × log10 x log10 x = ln x × log10 e = 0.434294 × ln x e = 2.718281828

APPENDIX 5 VECTOR IDENTITIES

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Až(B × C) = Bž(C × A) = Cž(A × B) A×(B × C) = B(AžC) − C(AžB) ∇(φ1 φ2 ) = φ1 ∇φ2 + φ2 ∇φ1 ∇ ž(A × B) = Bž(∇ × A) − Až(∇ × B) ∇×φA = ∇φ×A + φ(∇ × A) ∇ žφA = ∇φžA + φ(∇ žA) ∇ ž(∇ × A) = 0 ∇×(∇φ) = 0 ∇ × ∇ × A = ∇(∇ žA) − ∇ 2 A Stokes’s Theorem. The circulation around a simple closed curve is equal to the integral over any simple surface spanning the curve, of the normal component of the curl, the positive sense on the curve being counterclockwise as seen from the side of the surface toward which the positive normal points. Mathematically, Aždl = (∇ × A)žds C

S

11. Divergence Theorem (also known as Gauss’s theorem). The integral of the divergence of a vector field over a region of space is equal to the Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

593

594

VECTOR IDENTITIES

integral over the surface of that region of the component of the field in the direction of the outward directed normal to the surface. Mathematically, ž (∇ A)dv = Ažds V

S

12. Helmholtz’s Theorem. A vector is specified uniquely if its divergence and curl are given within a region and its normal component is given over the boundary. Cartesian Coordinates ∇f = xˆ ∇2 f =

∂f ∂f ∂f + yˆ + zˆ ∂x ∂y ∂z

∂ 2f ∂ 2f ∂ 2f + 2 + 2 2 ∂x ∂y ∂z

∂Ax ∂Ay ∂Az + + ∂x ∂y ∂z xˆ yˆ zˆ ∂ ∂ ∂ ∇ × A = ∂x ∂y ∂z Ax Ay Az ∇ žA =

Cylindrical Coordinates ∂f 1 ∂f ∂f + φˆ + zˆ ∂ρ ρ ∂φ ∂z 1 ∂ ∂ 2f ∂f 1 ∂ 2f ∇2f = ρ + 2 2 + 2 ρ ∂ρ ∂ρ ρ ∂φ ∂z ∇f = ρˆ

1 ∂(ρAρ ) + ρ ∂ρ ρˆ ρφˆ ∂ ∂ ∇ × A = ∂ρ ∂φ Aρ ρAφ ∇ žA =

1 ∂Aφ ∂Az + ρ ∂φ ∂z zˆ ∂ ∂z Az

Spherical Coordinates ∂f 1 ∂f 1 ∂f + θˆ + φˆ ∂r r ∂θ r sin θ ∂φ 1 ∂ ∂ ∂f 1 ∂ 2f 2 2 ∂f ∇ f = 2 sin θ r + sin θ + r sin θ ∂r ∂r ∂θ ∂θ sin θ ∂φ2 ∇f = rˆ

VECTOR IDENTITIES

1 ∂ ∂Aφ ∂ 2 ∇A= 2 sin θ (r Ar ) + r (sin θAθ ) + r r sin θ ∂r ∂θ ∂φ rˆ r θˆ r sin θφˆ 1 ∂ ∂ ∂ ∇×A= 2 r sin θ ∂r ∂θ ∂φ Ar rAθ r sin θAφ ž

595

APPENDIX 6 SOME USEFUL NETWORK TRANSFORMATIONS

This appendix contains transformations of networks that are commonly encountered. These transformations can save time of analysis and design. Figure A6.1 shows a series-connected Rs –Ls circuit transformed to a parallel-connected Rp –Lp circuit. Resistors in the two circuits may represent loss associated with the inductor. These two circuits are equivalent if their impedance (as well as their admittance) values are equal. This equality gives Rp = Rs +

(ωLs )2 Rs

(A6.1)

Lp = Ls +

Rs2 ω2 Ls

(A6.2)

and

Transforming the other way gives Rs =

Rp (ωLp )2 Rp2 + (ωLp )2

(A6.3)

Ls =

Lp (Rp )2 Rp2 + (ωLp )2

(A6.4)

and

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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SOME USEFUL NETWORK TRANSFORMATIONS

597

Ls Lp

Rp

Rs

Figure A6.1 Transformation of a series R–L to a shunt R–L circuit, or vice versa.

Cs Cp

Rs

Figure A6.2

Rp

Transformation of a series R–C to a shunt R–C circuit, or vice versa.

Similarly, a series Rs –Cs circuit can be transformed to a parallel Rp –Cp circuit, or vice versa, as illustrated in Figure A6.2. By equating their impedance (or admittance) expression, following relations may be found: Rp =

1 + (ωCs Rs )2 Rs (ωCs )2

(A6.5)

Cp =

Cs 1 + (ωCs Rs )2

(A6.6)

Rs =

Rp 1 + (ωCp Rp )2

(A6.7)

Cs =

1 + (ωCp Rp )2 ω2 Cp Rp2

(A6.8)

and

Transforming other way gives

and

The inductor-tapped circuit shown in Figure A6.3 may be transformed to a parallel Rp –Lp circuit. This type of transformation facilitates the analysis of a Hartley oscillator. Again, the impedance of an inductor-tapped circuit must be equal to that of a parallel Rp –Lp circuit. Therefore, Rp =

(ωL1 L2 )2 + R 2 (L1 + L2 )2 RL22

(A6.9)

Lp =

(ωL1 L2 )2 + R 2 (L1 + L2 )2 ω2 L1 L22 + R 2 (L1 + L2 )

(A6.10)

and

598

SOME USEFUL NETWORK TRANSFORMATIONS

L1 Lp

L2

Rp

R

Figure A6.3

Transformation of an inductor-tapped circuit.

C1 Cp

C2

Rp

R

Figure A6.4

Transformation of a capacitor-tapped circuit.

Similarly, a capacitor-tapped circuit can be transformed to a shunt Rp –Cp circuit as illustrated in Figure A6.4. In this case, the following relations must hold if the two circuits are equal: Rp =

1 + ω2 R 2 (C1 + C2 )2 ω2 RC12

(A6.11)

Cp =

C1 [1 + ω2 R 2 C2 (C1 + C2 )] 1 + ω2 R 2 (C1 + C2 )2

(A6.12)

and

APPENDIX 7 PROPERTIES OF SOME MATERIALS

TABLE A7.1 Conductivity of Some Materials Material Aluminum Brass Bronze Copper Gold Iron Nichrome Nickel Silver Stainless steel

Conductivity (S/m) 3.54 1.5 1.0 5.813 4.1 1.04 0.09 1.15 6.12 0.11

× × × × × × × × × ×

107 107 107 107 107 107 107 107 107 107

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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600

PROPERTIES OF SOME MATERIALS

TABLE A7.2 Dielectric Constant and Loss Tangent of Some Materials at 3 GHz Material Alumina Glass (Pyrex) Mica (ruby) Nylon (610) Polystyrene Plexiglas Quartz (fused) Rexolite (1422) Styrofoam (103.7) Teflon Water (distilled)

Dielectric Constant, εr

Loss Tangent, tan δ

9.6 4.82 5.4 2.84 2.55 2.60 3.8 2.54 1.03 2.1 77

0.0001 0.0054 0.0003 0.012 0.0003 0.0057 0.00006 0.00048 0.0001 0.00015 0.157

APPENDIX 8 COMMON ABBREVIATIONS

AAU ABCMOS ABR ACI ACTS ADC ADPCM ADSL AFC AGC AM AMPS ANSI ARDIS ASCII ASIC ASK ATM AWGN

analog audio advanced bipolar CMOS available bit rate adjacent channel interference advanced communication technology satellite analog-to-digital converter adaptive differential pulse code modulation asymmetric digital subscriber line automatic frequency control automatic gain control amplitude modulation advanced mobile phone service American National Standards Institute advanced radio data information service American Standard Code for Information Interchange application-specific integrated circuit amplitude-shift keying asynchronous transfer mode additive white Gaussian noise

B-CDMA BER BIFET

broadcast CDMA bit error rate bipolar field-effect transistor

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

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602

BIMOS BIOS BJT BONDING BPSK

COMMON ABBREVIATIONS

bipolar metal-oxide semiconductor basic input–output system bipolar junction transistor bandwidth on demand interoperability biphase-shift keying

CAD computer-aided design CBR continuous bit rate CCD charge-coupled device CCIR International Radio Consultative Committee CDMA code-division multiple access CDPD cellular digital packet data CIR carrier-over-interference ratio CMOS complementary metal-oxide semiconductor CNR carrier-over-noise ratio COB chip on board Codec coder–decoder CRC cyclic redundancy check CSDN circuit-switched digital network CSMA/CD carrier sense multiple access with collision detection CT cordless telephone CW continuous wave DAB DAC DAS DBS DCE DCP DCPSK DCS-1800 DDS DECT DFT DIP DMR DPCM DPSK DQPSK DRAM DRO DSMA DSO DSP DUT DWT

digital audio broadcast digital-to-analog converter data acquisition system direct broadcast satellite data communication equipment data communication protocol differentially coherent phase-shift keying digital communication system 1800 direct digital synthesis digital European cordless telecommunications discrete Fourier transforms dual-in-line package digital mobile radio differential pulse code modulation differential phase-shift keying differential quadrature phase-shift keying dynamic random access memory dielectric resonator oscillator digital sensed multiple access digital sampling oscilloscope digital signal processing device under test discrete wavelet transforms

COMMON ABBREVIATIONS

603

EBS EDA EIA EIRP EMC EMI ERC ESD ESDI ETN ETSI

emergency broadcast system electronic design automation Electronics Industries Association effective isotropic radiated power electromagnetic compatibility electromagnetic interference European Radio Commission electrostatic discharge enhanced small device interface electronic tandem network European Telecommunication Standard Institute

FCC FDD FDDI FDMA FET FFT FH FIFO FIR FM FPGA FSK

Federal Communications Commission frequency-division duplex fiber-distributed data interface frequency-division multiple access field-effect transistor fast Fourier transforms frequency hopping first-in first-out system finite impulse response filter frequency modulation field-programmable gate array frequency-shift keying

GEO GFSK GIGO GMSK GPIB GPS GSM GUI

geosynchronous satellite Gaussian frequency shift keying garbage in, garbage out Gaussian minimum shift keying general-purpose interface bus global positioning system (Groupe Special Mobile) global system for mobile communication graphical user interface

HBT HDTV HEMT HF HIPERLAN

heterojunction bipolar transistor high-definition television high-electron mobility transistor high frequency high-performance radio LAN

I and Q IC IDE IF IGBT IIT

in-phase and quadrature phase integrated circuits integrated device electronics intermediate frequency insulated gate bipolar transistor infinite impulse response filter

604

COMMON ABBREVIATIONS

IMD IMPATT INTELSAT IP IP IP2 IP3 ISDN ISI ISM ITU

intermodulation distortion impact ionization avalanche transit time diode International Telecommunication Satellite Consortium intermodulation product Internet protocol second-order intercept third-order intercept integrated services digital network intersymbol interference industrial, scientific, and medical bands International Telecommunication Union

JDC JFET JPEG

Japanese digital cellular standard junction field-effect transistor Joint Photographic Experts Group

LAN LATA LCD LED LEOs LHCP LNA LO LOS LUT

local area network local access and transport area liquid-crystal display light-emitting diode low-Earth-orbit satellite left-hand circular polarization low-noise amplifier local oscillator line of sight lookup table

MAC MAN MAU MBE MDS MDSL MEOs MESFET MIDI MIPS MLC MMI MMIC MMSI MODFET MOSFET MPEG MPP MPSK

medium access control metropolitan area network multistation access unit molecular beam epitaxy minimum detectable signal medium-bit-rate digital subscriber line medium-Earth-orbit satellite metal-semiconductor field-effect transistor musical instrument digital interface million instructions per second multilayer ceramics man–machine interface monolithic microwave integrated circuit multimode stereoscopic imaging modulation doped field-effect transistor metal-oxide-semiconductor field-effect transistor Motion Picture Experts Group massively parallel processing minimum phase-shift keying

COMMON ABBREVIATIONS

MSK MSP MSS MTI MTSO MUX

minimum shift keying mixed signal processor mobile satellite service moving-target indicator mobile telephone switching office multiplexing

NADC NF NMT NTSC NTT

North American Digital Cellular/Cordless noise figure Nordic Mobile Telephone National Television Standards Committee Nippon Telephone and Telegraph

OFDM OOK OOP

orthogonal frequency-division multiplexing on–off keying object-oriented programming

PAD PAL PAM PAMR PBX PCB PCM PCMCIA PCN PCS PDA PDC PDIP PGA PHP PHS PID PIN PLC PLL PM PMR POTS PPGA PQFP PSK PSTN PVC

packet assembler and disassembler phase alternating line pulse amplitude modulation public access mobile radio private branch exchange printed circuit board pulse code modulation Personal Computer Memory Card International Association personal communication network personal communication service personal digital assistant personal digital cellular plastic dual-in-line package programmable gain amplifier; also pin grid array personal handy phone personal handyphone system (formerly PHP) proportional integrating differentiating p-type, intrinsic, n-type diode programmable logic controller phase-locked loop pulse modulation private mobile radio plain old telephone service plastic pin grid array plastic quad flat pack phase-shift keying public switched telephone network permanent virtual connection

605

606

COMMON ABBREVIATIONS

PWB PWM

printed wire board pulse width modulation

QAM QCELP QDM QMFSK QPSK

quadrature amplitude modulation Qualcomm coded excited linear predictive coding quadrature demodulator quadrature modulation frequency-shift keying quadrature phase-shift keying

RADAR RAID RBDS RCS RES RF RFI RHCP RMS RPE-LTP RTD RTMS

radio detection and ranging redundant array of independent disks radio broadcast data system radar cross section Radio Expert Systems Group radio frequency radio-frequency interference right-handed circular polarization root mean square regular pulse excitation long-term predictor resistive thermal detector radio telephone mobile system

SAW SCSI SDH SDIP SDN SFDR SINAD SIP SLICs SMD SMR SNR SOI SOIC SONET SOT SPLIC SQPSK SQUID SSB STP SVD SWR

surface acoustic wave small computer systems interface synchronous digital hierarchy shrink dual-in-line package switched digital network spur-free dynamic range signal-over-noise and distortion single-in-line package standard linear integrated circuits surface-mount device specialized mobile radio signal-to-noise ratio silicon on insulator small outline integrated circuit synchronous optical network small outline transistor special-purpose linear integrated circuit staggered quadriphase-shift keying superconducting quantum interface single sideband shielded twisted pair simultaneous voice and data standing wave ratio

COMMON ABBREVIATIONS

607

TACS TCP T-DAB TDD TDMA TDR TE TEM TETRA THD+N TIA TM TNC TSAPI TTL TVG TVRO TVS TWTA Tx

total access communication system transmission control protocol terrestrial-digital audio broadcasting time-division duplex time-division multiple access time-domain reflectometry transverse electric transverse electromagnetic wave trans-European trunked radio system total harmonic distortion plus noise Telecommunications Industry Association transverse magnetic threaded Neil-Councilman connector telephony services–oriented application programming interface transistor–transistor logic time-variable gain TV receive-only transient voltage suppression traveling wave tube amplifier transmitter

UART UDPC UHF UNC UNI UPS USART USB UTC UTP

universal asynchronous receiver/transmitter universal digital personal communications ultrahigh frequency universal control network user-to-network interface uninterruptible power supply universal synchronous–asynchronous receiver transmitter universal serial bus universal time code unshielded twisted pair

VCO VDT VHDL VHF VLF VLSI VPSK VSAT VSELP VSWR VXCO

voltage-controlled oscillator video display terminal very high definition language very high frequency very low frequency very-large-scale integration variable phase-shift keying very small aperture (satellite ground) terminal vector sum excited linear predictive coding voltage standing wave ratio voltage-controlled crystal oscillator

WAN WATS

wide area network wide area telecommunication network

608

WDM WER WFAU WLAN WLL WORM WPABX

COMMON ABBREVIATIONS

wavelength-division multiplexing word error rate wireless fixed access unit wireless local area network wireless local loop write once, read many wireless private branch exchange

APPENDIX 9 PHYSICAL CONSTANTS

Permittivity of free space, ε0 Permeability of free space, µ0 Impedance of free space, η0 Velocity of light in free space, c Charge of electron, qe Mass of electron, me Boltzmann’s constant, k Planck’s constant, h

8.8542 × 10−12 F/m 4π × 10−7 H/m 376.7 2.997925 × 108 m/s 1.60210 × 10−19 C 9.1091 × 10−31 kg 1.38 × 10−23 J/K 6.6256 × 10−34 J·s

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

609

INDEX

1/f noise, 34 Acquisition of lock, 509 Admittance matrix, 289 Admittance parameters, 289 AM, see Amplitude modulation Ampere’s law, 105 Amplitude modulation, 544 DSB, 546 modulation index, 544 SSB, 546 suppressed carrier AM, 546 Antenna efficiency, 21 Antennas, 18 Cassegrain reflector, 18 directive gain, 19 electric dipole, 18 isotropic antenna, 19 lens, 18 power gain, 19 pyramidal horn, 18 Attenuation constant, 62, 116 Available power gain, 411 Balanced amplifier, 468 Bandpass filter, 371 Bandstop filter, 375 Barkhausen criterion, 480

Bessel function, 143, 180, 556 Binomial filter, 354 Binomial transformers, 244 Bisected π-section, 346 Black-body radiation law, 34 Bode–Fano constraints, 280 Broadband amplifiers, 466 Butterworth filter, 354 C3 , 7 Carson rule bandwidth, 556 Cellular telephones, 17 Chain matrix, 298 Chain scattering parameters, 325 Characteristic impedance, 59 experimental determination of, 75 Chebyshev filter, 355 Chebyshev polynomials, 250, 356 Chebyshev transformers, 248, 255 Clapp oscillator, 488 Closed-loop gain, 480 Coaxial line, 582 Coefficient of coupling, 166 Coherent detector, 553 Colpitts oscillator, 484 Complex permittivity, 112 Composite filters, 346

Radio-Frequency and Microwave Communication Circuits: Analysis and Design, Second Edition, By Devendra K. Misra ISBN 0-471-47873-3 Copyright 2004 John Wiley & Sons, Inc.

611

612 Conductivity, 59, 112 Conservation of charge, 107 Constant gain circles, 439 Constant noise figure circles, 457 Constant-k, 337 Constitutive relations, 111 Cordless telephones, 17 Critically damped, 153 Current reflection coefficient, 73 Current sensitivity, 551 Cutoff frequency, 337, 338, 354, 360, 369 Cylindrical cavity, 179 Cylindrical waveguide, 137, 142 Damping ratio, 153 dB, 580 dBc, 581 dBW, 580 Dibit, 9 Directivity, 19, 27 Distortion, 33, 45 Distortionless line, 75 Distributed amplifiers, 468 Distributed elements, 58 Doppler radar, 11, 32 Divergence theorem, 593 Double-balanced mixer, 563 Double-stub tuner, 202 DR, see Dielectric resonators Dynamic conductance, 550 Dynamic range, 5, 11, 47 EIRP, 24 Electric charge density, 105 Electric current density, 105 Electric field intensity, 104, 112 Electric flux density, 105, 111 Electric scalar potential, 128 Electric vector potential, 129 Electrical susceptibility, 111 Equation of continuity, 107 Faraday’s law of induction, 105 FDM, 543 FET mixer, 571 Flicker noise, 34 FM, see Frequency modulation FM detector, 558 Fractional bandwidth, 236 Frequency bands, 2, 3 Frequency converter, 547 Frequency division multiplexing, 543 Frequency modulation, 2, 9, 543, 555 modulation index, 555

INDEX Frequency scaling, 360 Frequency-shift keying, 9 Friis transmission formula, 26 Gain compression, 48 Geosynchronous, 13 Gilbert cell, 575 Global positioning system, 15 Group velocity, 65 Gunn diodes, 520 Half-power beam width, 20 Hankel functions, 144 Hartley oscillator, 484 HBT, 422 Helmholtz’s equations, 62 Helmholtz’s theorem, 128 HEMT, 422 High-pass filter, 338, 369 Hybrid parameters, 296 Image impedance, 334 Image parameter method, 334 IMD, 569 IMPATT diode, 520 Impedance matrix, 284 Impedance parameters, 284 Impedance scaling, 361 Incident wave, 64, 123 Indirect frequency synthesizer, 516 Input node, 396 Input reflection coefficient, 68 Input stability circle, 427 Insertion loss, 72 Insertion loss method, 353 Intercept point, 50 Intermediate frequency, 12 Intermodulation distortion, 45, 567 Intermodulation distortion ratio, 49 Intrinsic impedance, 115 IP, see Intercept point IR, 2 Isotropic antenna, 19 Johnson noise, 33 Klopfenstein taper, 276 Kuroda’s identities, 384 Leontovich impedance boundary condition, 121 LEOS, 15, 16 Line-of-sight propagation, 5 Load reflection coefficient, 70, 80

INDEX Lock-in range, 510 Loop, 398, 480 Loop gain, 398, 480 Loop-filters, 500 Lorentz condition, 128, 130 Loss-tangent, 112 Low-pass filter, 337, 354 Low-pass filter synthesis, 358 Low-pass ladder network prototype, 358 Magnetic charge, 126 Magnetic current, 126 Magnetic field intensity, 104 Magnetic flux density in, 105 Magnetic scalar potential, 129 Magnetic susceptibility, 112 Magnetic vector potential, 127 Magnetization density, 112 Maximally flat filter, 354 Maximally flat low-pass filters, 355 Maxwell’s equations, 109, 110 m-derived filter, 341 MEOS, 15 Microstrip line, 585 Microwave transistor oscillator, 523 Minimum detectable signal, 44 Minkowski inequality, 424 Mixer, 9, 12 Mixers: conversion loss, 565 diode ring, 567 dual gate FET, 574 FET mixers, 571 intermodulation distortion, 569 single diode, 548 switching type, 559 Negative resistance, 489 Neper, 581 Neumann function, 143 Noise, 33 Noise factor, 37 Noise figure, 38 Noise figure of a cascaded system, 40 Noise figure parameter, 458 Noise in two-port networks, 39 Noise temperature, 35 Nonredundant circuit synthesis, 380 Normalized image impedance, 344 Normalized input impedance, 68 Notch-filter, 333 Nyquist criterion, 480 Nyquist noise, 33

On-off keying, 9 Operating power gain, 414 Output node, 396 Output stability circle, 426 Overdamped, 153 Parallel-plate waveguide, 133 Path, 398 Path gain, 398 Personal communication networks, 16 Personal handy phone, 17 Phase constant, 62, 114 Phase detector, 498 Phase modulation, 9 Phase velocity, 65, 115 Phase-locked loop, 497 Phase-shift keying, 9 Pierce oscillator, 494 Pink noise, 34 Polarization, 23 Polarization density, 111 Power density, 19 Power flow, 113 Power loss ratio, 353 Poynting’s vector, 113 Propagation constant, 62, 116 experimental determination of, 75 Pull-in range, 510 Pulling figure, 494 Q, see Quality factor Quadrature phase-shift keying, 9 Quality factor, 155 Quarter-wavelength transformer, 70 Radar, 4 Radar cross-sections, 29 Radar range equation, 29 Radiation efficiency, 22 Radiation field, 19 Radiation intensity, 19 Radiation pattern, 20 Radio frequency detector, 551 Radio frequency identification, 2 Rayleigh–Jeans approximation, 34 Reactive L-section matching, 209 Receiver, 8 Rectangular cavity, 177 Redundant filter synthesis, 380 Reflected wave, 64 Reflection coefficient, 68 Remote sensing, 7 Repeater, 12 Resistive L-section matching, 207

613

614

INDEX

Resonant frequency, 179 Return-loss, 72 Reverse-saturation current, 549 Richard’s transformation, 384 Sampling theorem, 272 Satellite communication, 13 Scattering matrix, 306 Scattering parameters, 306 shifting of reference planes, 311 Scattering transfer parameters, 325 Second harmonic distortion, 48 Self-loop, 398 Semirigid coaxial lines, 589 Shot noise, 33 Shunt matching element, 190 Shunt stub matching, 190 Signal flow graph, 392 branches, 392 junction points, 392 Mason’s rule, 399 nodes, 663 rules of reduction, 398 Signal-to-noise ratio, 37 Skin depth, 115 Simultaneous conjugate matching, 432 Single-balanced mixer, 560 Single-diode mixer circuit, 548 Small-signal equivalent circuit of a BJT, 469 Small-signal equivalent circuit of a MESFET, 471 Small-signal equivalent circuit of a MOSFET, 470 Smith chart, 85 Space loss, 25 Spurious-free dynamic range, 51 SSB, 546 Stability circle, 427 Stokes’ theorem, 593 Strip line, 583 Stub, 189 Switching-type mixers, 559 Synchronization range, 507 Terrestrial communication, 12 Three-port S-parameter description of the transistor, 530 Time-harmonic fields, 110

Time-period, 64 Tracking range, 507 Transducer power gain, 412 Transmission line: attenuation constant, 62 characteristic impedance, 59 distributed network model, 61 line parameters, 58 phase constant, 62 propagation constant, 62 Transmission line equations, 61 Transmission lines, 4 Transmission parameters, 298 Transmitter, 8 Transponder, 14 Traveling wave amplifiers, 468 Traveling wave tube, 14 Tuning sensitivity, 496 UHF, 3 Undamped natural frequency, 153 Underdamped, 153 Unilateral figure of merit, 437 Unilateral transducer power gain, 413 Varactors, 495 VCO, see Voltage-controlled oscillator Vector wave equation, 114 VHF, 3 Voltage-controlled oscillator, 496, 497 Voltage reflection coefficient, 72 Voltage sensitivity, 552 Voltage standing wave, 82 VSAT, 15 VSWR, 82 Wavelength, 64 Wave number, 114 Wire antennas, 18 electric dipole, 18 loop antennas, 18 Wireless communication, 8, 16 Y -factor method, 36 Zero-biased diode detector, 552 ZY -Smith chart, 219

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